A stone is thrown horizontally with an initial speed of 10m/s from the edge of the cliff. A stop watch measures the stone’s trajectory with time from the top of the hill to the bottom to be 6.7s. What is the height of the cliff?

Answer:

At a distance of 100 m from the source the intensity will be 40 dB.

Explanation:

Sound intensity is the acoustic power transferred by a sound wave per unit area normal to the direction of propagation.

The sound intensity depends on the power of the sound source, where the higher the power the greater the intensity, the distance to the sound source, the greater the distance being the lower the intensity, and the nature of the transmission medium.

The conversion between intensity and decibels corresponds to:

[tex]L=10*log\frac{I}{I0}[/tex]

where I0 = 10⁻¹² W/m² and corresponds to a level of 0 decibels therefore.

In this case, you can apply the following relationship between two intensities and distance, considering that the intensity of the sound level decreases with distance:

[tex]L1 - L2=10*log\frac{I1}{I0} - 10*log\frac{I2}{I0}[/tex]

[tex]L1 - L2=10*(log\frac{I1}{I0} - *log\frac{I2}{I0})[/tex]

[tex]L1 - L2=10*[log(\frac{I1}{I0}\frac{I0}{I2})][/tex]

[tex]L1 - L2=10*[log(\frac{I1}{I2})][/tex]

Being L1= 70 dB and L2= 40 dB

[tex]70 dB - 40 dB=10*[log(\frac{I1}{I2})][/tex]

[tex]30=10*[log(\frac{I1}{I2})][/tex]

[tex]\frac{30}{10} =log(\frac{I1}{I2})[/tex]

[tex]3=log(\frac{I1}{I2})[/tex]

[tex]10^{3} =\frac{I1}{I2}[/tex]

[tex]1,000=\frac{I1}{I2}[/tex]

The intensity is inversely proportional to the square of the distance to the source. The relationship between the intensities I1 and I2 at distances d1 and d2 respectively is:

[tex]\frac{I1}{I2} =\frac{d2^{2} }{d1^{2} }[/tex]

Then:

[tex]1,000=\frac{d2^{2} }{d1^{2} }[/tex]

Being d1= 10 m

[tex]1,000=\frac{d2^{2} }{10^{2} }[/tex]

[tex]1,000=\frac{d2^{2} }{100}[/tex]

1,000*100= d2²

10,000= d2²

√10,000= d2

100 m= d2

At a distance of 100 m from the source the intensity will be 40 dB.

Answer:

2.59 cm

Explanation:

The torque τ on a current carrying loop of wire is given by τ = NiABsinθ where N = number of turns of loop, i = current in loop, A = area of loop and B = magnetic field.

Now, given that τ = 0.0256 Nm, i = 1.10 A, B = 1.20 T,N = 30 and since the loop is tilted 15° off the x-axis and the magnetic field points in the negative y- direction, the angle between the normal to the loop and the magnetic field is thus 90° - 15° = 75°. So, θ = 75°.

We now find the area of the loop A from

τ = NiABsinθ

A = τ/NiBsinθ

substituting the values of the variables, we have

A = 0.0256 Nm/30 × 1.10 A × 1.20 T × sin75°

A = 0.0256 Nm/38.25

A = 6.69 × 10⁻⁴ m²

Since the loop is a square, with length of side L, its area A = L² and

L = √A

= √(6.69 × 10⁻⁴ m²)

= 2.59 × 10⁻² m

converting to cm, we have

L = 2.59 × 10⁻² m × 100 cm/m

L = 2.59 cm

So, the lengths of sides of the loop is 2.59 cm

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is [tex]u  =  7.73 \  m/s [/tex]

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      [tex]s = h+  u t +  0.5 gt^2[/tex]

Here s  is the distance covered by the bag at sea level which is zero

      [tex]0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2[/tex]

=>    [tex]0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2[/tex]

=>   [tex]u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}[/tex]

=>   [tex]u  =  7.73 \  m/s [/tex]

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

[tex]y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex] (Eq. 1)

Where:

[tex]y_{o}[/tex] - Initial height of the softball, measured in meters.

[tex]y[/tex] - Final height of the softball, measured in meters.

[tex]v_{o}[/tex] - Initial velocity of the softball, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

If we know that [tex]y = y_{o}[/tex], [tex]t = 3.56\,s[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the initial velocity of the softball is:

[tex]v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0[/tex]

[tex]3\cdot v_{o} -44.132\,m= 0[/tex]

[tex]v_{o} = 14.711\,\frac{m}{s}[/tex]

The initial velocity of the softball is 14.711 meters per second.

Answer:

vₓ = 53.6 km/h

vy = 12.4 km/h

Explanation:

       [tex]v_{x} = v_{o} * cos \theta = 55 km/h * cos 13 = 53.6 km/h[/tex]

       [tex]v_{y} = v_{o} * sin\theta = 55 km/h * sin 13 = 12.4 km/h[/tex]

Answer:

A) m[tex]_{T}[/tex] = 0.3025 * 0.0476 * 92.13 = 1.327 Ibm

B) T= 63.32°F

Explanation:

Given data:

1000 gallon tank currently contains 100.0 gallons of liquid toluene

and A gas saturated with toluene vapor at 85°F and 1 atm

A) Calculate quantity of toluene ( Ibm ) that will enter the atmosphere when the tank is filled

m[tex]_{T}[/tex] = [tex]n_{gas} * Y_{T} * M_{T}[/tex]

[tex]n_{gas}[/tex] (total mole of gas) = 0.3025 Ib-mole  ( calculated using : [tex]\frac{PV}{RT}[/tex] )

[tex]Y_{T}[/tex] (mole fraction of toluene) = 0.0476 ( calculated using [tex]\frac{P_{T} }{P}[/tex] )

M[tex]_{T}[/tex] = 92.13 Ibm/Ib-mole

therefore:  m[tex]_{T}[/tex] = 0.3025 * 0.0476 * 92.13 = 1.327 Ibm

B) using Antoine equation to solve for T

Antoine equation : [tex]log_{10} (P_{T} ) = A - \frac{B}{T+C}[/tex]

PT( partial pressure ) = 18.95 ( calculated using : [tex]y_{tb} * P[/tex] )

A = 6.95805

B = 1346.773

T = ?

C = 219.693

to calculate T make T the subject the subject of the equation

T + 219.693 = 1346.773 / 5.68044

∴ T = 17.40°C

convert T to Fahrenheit

T = 1.8 * 17.40 +32

  = 63.32°F

Answer:

v₃ = 0.786 m/s

Explanation:

Here, we will use the law of conservation of momentum, which states the following:

Total Momentum of System Before Collision =

Total Momentum of System After Collision

m₁u₁ + m₂u₂ + m₃u₃ = m₁v₁ + m₂v₂ + m₃v₃

where,

m₁ = mass of bullet = 6.13 g = 0.00613 kg

m₂ = mass of 1st block = 1233 g = 1.233 kg

m₃ = mass of 2nd block = 1646 g = 1.646 kg

u₁ = speed of first bullet before collision = 361 m/s

u₂ = speed of first block before collision = 0 m/s

u₃ = speed of 2nd block before collision = 0 m/s

v₁ = speed of bullet after collision

v₂ = speed of 1st block after collision = 0.741 m/s

v₃ = speed of 2nd block after collision = ?

Therefore,

(0.00613 kg)(361 m/s) + (1.233 kg)(0 m/s) + (1.646 kg)(0 m/s) = (0.00613 kg)(v₁) + (1.233 kg)(0.741 m/s) + (1.646 kg)(v₃)

2.2129 kg m/s + 0 kg m/s + 0 kg m/s - 0.9136 kg m/s = (0.00613 kg)(v₁) + (1.233 kg)(0.741 m/s) + (1.646 kg)(v₃)

1.2992 kg m/s = (0.00613 kg)(v₁) + (1.646 kg)(v₃)

since, the bullet is embedded in 2nd block after collision. Thus, there velocities will become same. (v₁ = v₃)

Therefore,

1.2992 kg m/s = (0.00613 kg)(v₃) + (1.646 kg)(v₃)

v₃ = (1.2992 kg m/s)/(1.6521 kg)

v₃ = 0.786 m/s

Answer:

I dont know sorry,I hope you find what are you looking for.

Answer:

The ratio of the displacement amplitudes of two sound waves is 1.16.

Explanation:

Given that,

Frequency = 5.0 kHz

Intensity level difference = 3.0 dB

We know that,

The sound intensity is inversely proportional to the square of distance.

[tex]I\propto\dfrac{1}{r^2}[/tex]

The sound intensity for first wave is

[tex]\beta_{1}=10\log\dfrac{I_{1}}{I_{0}}[/tex]...(I)

The sound intensity for second wave is

[tex]\beta_{2}=10\log\dfrac{I_{2}}{I_{0}}[/tex]...(II)

We need to calculate the ratio of intensity

From equation (I) and (II)

[tex]\beta_{2}-\beta_{1}=10\log\dfrac{I_{2}}{I_{0}}-10\log\dfrac{I_{1}}{I_{0}}[/tex]

[tex]\Delta \beta=10\log(\dfrac{I_{2}}{I_{1}})[/tex]

Put the value into the formula

[tex]3.0=10\log(\dfrac{I_{2}}{I_{1}})[/tex]

[tex]\dfrac{I_{2}}{I_{1}}=e^{\dfrac{3.0}{10}}[/tex]

[tex]\dfrac{I_{2}}{I_{1}}=1.34[/tex]

We need to calculate the ratio of the displacement

Using formula of displacement

[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{\dfrac{I_{2}}{I_{1}}}[/tex]

Put the value into the formula

[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{1.34}[/tex]

[tex]\dfrac{r_{1}}{r_{2}}=1.16[/tex]

Hence, The ratio of the displacement amplitudes of two sound waves is 1.16.

Answer:

The average velocity of the stream with a depth of 10 feet is greater.

(a) is correct option.

Explanation:

Given that,

Depth [tex]h_{1}= 10\ feet[/tex]

Depth [tex]h_{2}=6\ feet[/tex]

We need to calculate the average velocity of the stream

According to question,

[tex]h_{1} > h_{2}[/tex]

The velocity for first case,

[tex]v_{1}=u_{1}\dfrac{x_{1}}{h_{1}}[/tex]

[tex]\dfrac{v_{1}}{x_{1}}=\dfrac{u_{1}}{h_{1}}[/tex]

The velocity for second case,

[tex]v_{2}=u_{2}\dfrac{x_{2}}{h_{2}}[/tex]

[tex]\dfrac{v_{2}}{x_{2}}=\dfrac{u_{2}}{h_{2}}[/tex]

For the same velocity profile,

[tex]\dfrac{dv}{dx}=\dfrac{v_{1}}{x_{1}}=\dfrac{v_{2}}{x_{2}}[/tex]

Then,

[tex]\dfrac{u_{1}}{h_{1}}=\dfrac{u_{2}}{h_{2}}[/tex]

Put the value into the formula

[tex]\dfrac{u_{1}}{10}=\dfrac{u_{2}}{6}[/tex]

[tex]u_{1}=\dfrac{5}{3}u_{2}[/tex]

[tex]u_{1}=1.67u_{2}[/tex]

The velocity is [tex]u_{1} > u_{2}[/tex]

We need to calculate the average velocity for first case

Using formula of average velocity

[tex]v_{avg}_{1}=\dfrac{0+u_{1}}{2}[/tex]

Put the value into the formula

[tex]v_{avg}_{1}=\dfrac{0+u_{1}}{2}[/tex]

[tex]v_{avg}_{1}=\dfrac{u_{1}}{2}[/tex]

We need to calculate the average velocity for second case

Using formula of average velocity

[tex]v_{avg}_{2}=\dfrac{0+u_{2}}{2}[/tex]

Put the value into the formula

[tex]v_{avg}_{2}=\dfrac{0+u_{2}}{2}[/tex]

[tex]v_{avg}_{2}=\dfrac{u_{2}}{2}[/tex]

If [tex]u_{1} > u_{2}[/tex] then [tex]\dfrac{u_{1}}{2} >\dfrac{u_{2}}{2}[/tex]

So, we can say that the average velocity of the stream with a depth of 10 feet will be greater than the stream with a depth of 6 feet.

Hence, The average velocity of the stream with a depth of 10 feet is greater.

(a) is correct option.

Answer:

Taking a look at Newton's third law of motion which states "for every force exerted, their is an opposite force equal in magnitude and opposite in direction on the first force".

Similarly if a bullet had enough forces behind it to hurl someone through the air when they were hit, a similar force would act on the person holding the gun that fired the bullet.  

What we load into the gun is called a 'cartridge' Each piece is composed of four basic substance the casing, the bullet, the primer, and the powder.  

The primer explodes lighting the powder which causes a buildup of pressure behind the bullet. This powder can be used in rifle cartages because the bullet chamber is designed to withstand greater pressures.  

It is difficult in practice to measure the forces within a gun bagel, but the one easily measured parameter is the velocity with which the bullet exits muzzle velocity, therefore assuming that even if a handgun cartridge which generate enough momentum for the bullet to do this,  it is still nonsense on screen in Hollywood and video.  

Answer:

Explanation:

First thing we should note, is remember Avogadro's law, which states that "equal no of moles of gas occupy equal volume". And thus, the number of moles on both the sides of above are the same, this means that the number of particles that are colliding with wall due to which pressure will be present. Also, if the number of moles are the same, which they are, then the pressure will also be the same.

The Initial number of moles is(see first attachment), without ammonia.

At a point in the reaction, α is the amount or number of N2 that has reacted. Therefore, the total number of gas moles is get by is(see second attachment)

The first two terms represents a change in number of moles of reactant while the last one stands for the number of ammonia produced.

Answer:

At the moment, the truck's momentum (in kg · m/s)  is 2200 times as large as the truck's speed (in m/s).

This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).

If the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.

Explanation:

From the question,

The truck's momentum (in kg · m/s) is proportional to the truck's speed (in m/s).

Let the truck's momentum be P and the truck's speed be v,

Then  we can write that

P∝v

Then,

P = kv

Where k is the proportionality constant

From the question,

At some moment the truck's momentum is 50600 kg · m/s and the truck's speed is 23 m/s,

To determine how many times the truck's speed is as large as the truck's momentum at this moment, we will divide the truck's momentum by the speed, that is

50600 ÷ 23 = 2200

Hence, at the moment, the truck's momentum (in kg · m/s)  is 2200 times as large as the truck's speed (in m/s).

Since, dividing the truck's momentum by the truck's speed gives the proportionality constant k (that is, P/v = k), then

This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).

From

P = kv

Then, k = P/v

At a moment, P = 50600 kg · m/s and v = 23 m/s

∴ k = 50600 kg · m/s ÷ 23 m/s = 2200 kg

k = 2200 kg

To determine the truck's momentum if the truck is traveling at 16 m/s

From

P = kv

k = 2200 kg

v = 16 m/s

∴ P = 2200 kg × 16 m/s

P = 35200 kg · m/s

Hence, if the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.

Answer:

102 m

Explanation:

Given that It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

Let the stopping distance be equal to S.

According to the definition of speed,

Speed = distance / time.

make time the subject of the formula

Time = distance / speed

then, the equivalent time is:

48.96 / 12 = S / 25

Cross multiply

12S = 48.96 x 25

12S = 1224

S = 1224 / 12

S = 102 m

Therefore, the stopping distance is 102 m

Answer:

The rate the wheel will process about the vertical is 2.86 RPM

Explanation:

Given;

radius of the bicycle wheel, R = 0.36 m

mass of the wheel, m = 3.2 kg

angular velocity, ω = 4 rev/s

The rate at which the wheel will process about the vertical is given by;

Ф = τ/L

Where;

τ is the torque

L is the angular momentum

τ = Fr

τ = mgr = 3.2 x 9.8 x 0.1 = 3.126 N.m

L = Iω =  MR²ω

L = 3.2 x (0.36)² x (4 x 2π)

L = 10.4244 kg.m²/s

Ф = τ/L

Ф = (3.126) / (10.4244)

Ф = 0.29987 rad/s

Ф = 0.29987 rad/s  x (60 / 2π)

Ф = 2.86 RPM

Therefore, the rate the wheel will process about the vertical is 2.86 RPM

Answer:

a) 1.22*10^-18 N in the positive z direction

b) 1.65*10^-19 N in the negative z direction

c) (6.94*10^-19 N) in the positive x direction + (5.30*10^-19 N) in the positive z direction

Explanation:

See attachment for calculations

(a) The electromagnetic force on the proton is 1.224 × 10⁻¹⁸ [tex]\hat k[/tex] N

(b) The force is 1.65 × 10⁻¹⁹ [tex](-\hat k)[/tex] N

(c) The force is (6.94 [tex]\hat i[/tex] + 5.29 [tex]\hat k[/tex]) × 10⁻¹⁹ N

Given a proton moving in the positive y-direction with a speed of :

v = 1680 m/s [tex]\hat j[/tex]

The magnetic field is in the negative x-direction with magnitude:

B = 1.97 mT [tex](-\hat i)[/tex]

(a) Electric field applied in positive z-direction :

E = 4.34 V/m [tex]\hat k[/tex]

The net force on the proton is iven by:

F = q (E + v×B)

where q is the charge on proton, given by:

q = 1.6×10⁻¹⁹ C

So,

F = 1.6×10⁻¹⁹( 4.34 [tex]\hat k[/tex] + 1680 [tex]\hat j[/tex] × 1.97×10⁻³ [tex](-\hat i)[/tex] )

F = 1.6×10⁻¹⁹ ( 4.34 [tex]\hat k[/tex] + 3.309 [tex]\hat k[/tex])

F = 1.224 × 10⁻¹⁸ [tex]\hat k[/tex] N

(b) Electric field applied in negative z-direction :

E = 4.34 V/m [tex](-\hat k)[/tex]

The net force on the proton is iven by:

F = q (E + v×B)

where q is the charge on proton, given by:

q = 1.6×10⁻¹⁹ C

So,

F = 1.6×10⁻¹⁹( 4.34 [tex](-\hat k)[/tex] + 1680 [tex]\hat j[/tex] × 1.97×10⁻³ [tex](-\hat i)[/tex] )

F = 1.6×10⁻¹⁹ ( 4.34 [tex](-\hat k)[/tex] + 3.309 [tex]\hat k[/tex])

F = 1.65 × 10⁻¹⁹ [tex](-\hat k)[/tex] N

(c) Electric field applied in positive x-direction :

E = 4.34 V/m [tex]\hat i[/tex]

The net force on the proton is iven by:

F = q (E + v×B)

where q is the charge on proton, given by:

q = 1.6×10⁻¹⁹ C

So,

F = 1.6×10⁻¹⁹( 4.34 [tex]\hat i[/tex] + 1680 [tex]\hat j[/tex] × 1.97×10⁻³ [tex](-\hat i)[/tex] )

F = 1.6×10⁻¹⁹ ( 4.34 [tex]\hat i[/tex] + 3.309 [tex]\hat k[/tex])

F = (6.94 [tex]\hat i[/tex] + 5.29 [tex]\hat k[/tex]) × 10⁻¹⁹ N

Learn more about electromagnetic force:

https://brainly.com/question/807785?referrer=searchResults

Answer:

Convection

Explanation:

Answer:

C. Convection

Explanation:

I just did it

the answer is 5 m up AKA "B"

Answer:

W = 0.012 J

Explanation:

For this exercise let's use Hooke's law to find the spring constant

         F = K Δx

         K = F / Δx

         K = 3 / (0.16 - 0.11)

         K = 60 N / m

Work is defined by

         W = F. x = F x cos θ

in this case the force and the displacement go in the same direction therefore the angle is zero and the cosine is equal to 1

         W = ∫ F dx        

         W = k ∫ x dx

we integrate

         W = k x² / 2

          W = ½ k x²

let's calculate

         W = ½ 60 (0.19 -0.17)²

         W = 0.012 J

Answer:

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

From the question

volume = 56 ml

density = 1.24 g/ml

We have

mass = 56 × 1.24

We have the final answer as

Hope this helps you

Answer:

40.0⁰

Explanation:

The formula for calculating the magnetic flux is expressed as:

[tex]\phi = BAcos\theta[/tex] where:

[tex]\phi[/tex] is the magnetic flux

B is the magnetic field

A is the cross sectional area

[tex]\theta[/tex] is the angle that the normal to the plane of the loop make with the direction of the magnetic field.

Given

A = 0.250m²

B = 0.020T

[tex]\phi[/tex] = 3.83 × 10⁻³T· m²

3.83 × 10⁻³ = 0.020*0.250cosθ

3.83 × 10⁻³ = 0.005cosθ

cosθ = 0.00383/0.005

cosθ = 0.766

θ = cos⁻¹0.766

θ = 40.0⁰

Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰

Answer:

10-16= -4 so, 1.0x-4 is -4. So the answer is -4.

Explanation:

Complete question is;

The gravitational force acting on a lead ball is much larger than that acting on a wooden ball of the same size. Which statement is true about when they are dropped?

A) The acceleration due to gravity is the same for both balls.

B) The lead ball’s acceleration due to gravity is larger than the wooden ball’s.

C) The lead ball’s acceleration due to gravity is less than the wooden ball’s.

Answer:

Option A) The acceleration due to gravity is the same for both balls.

Explanation:

Density of Lead is 11.34 g/cm³ while density of hard wood is around 1.3 g/cm³

Thus, it means lead is the heavier object when both have same masses.

We are told that the wood and lead both have same size but from the density given above, it's clear that the mass of the lead ball is more than that of the wooden ball and thus, the force from would be more for the lead ball than the wooden ball.

Now, If we were to measure the position of both balls when falling, we will observe that they do not fall fall with a constant speed. This is because they fall with a constant acceleration and as such the speed increases.

Therefore both balls will hit the ground at the same time because they both start from a position of rest and both will have the same acceleration.

Answer:

The car appears to have a constant, positive acceleration for most of the video clip.

Answer:

F equals 3 N and the object remains stationary. (second option in the list)

Explanation:

For sure to cancel acting forces, F must be 3N pointing up. But with regards to the object stationary or not, the question is tricky. We could have a ZERO net force applied, and the object moving at constant speed, which could still verify Newton's Laws. But considering the first answer option that refers to vertical motion upward where the object could be gaining potential energy, the most accurate response is that the force F has to be 3 N pointing up to make the object in equilibrium, and no motion in the vertical axis.

Answer: F equals 3 N and the object remains stationary.

Explanation:

Answer:

(a) The net force is 80.394 N

    The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, [tex]F_y = Fsin \theta = 250sin45 = 176.78 \ N[/tex]

Applied force in x-direction, [tex]F_x = Fcos \theta = 250cos45 = 176.78 \ N[/tex]

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

[tex]F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N[/tex]

Apply Newton's second law of  motion;

F = ma

[tex]a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2[/tex]

(b) the velocity of the crate after 5.0 s

[tex]F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s[/tex]

Answer:

the answer is 1 and 4

Explanation

Plato users

For the potential energy, statement 1 and statement 4 are correct.

The potential energy of the object the energy of the object in its steady position. When the object is at rest, the energy of the object in that condition is called potential energy.

Let us consider an electron having charge [tex]e[/tex] is moving the distance [tex]d[/tex] in uniform electric field E.

Its potential energy can be written as,

[tex]P = eEd[/tex]

Where P is the potential energy and E is the electric field.

Hence, the potential energy of the electron is associated with the electric field.

The electric force can be written as,

[tex]F =eE[/tex]

Where [tex]F[/tex] is the electric force, [tex]E[/tex] is the electric field and [tex]e[/tex] is the charge on the electron.

So, the potential energy can be written as,

[tex]P=Fd[/tex]

Hence, the potential energy is related to electric force.

For more information, follow the link given below.

https://brainly.com/question/1413008.

Answer:

7 Newton's East

Explanation:

when the force is going in the same direction in this case east, you add the forces.

Answer:

T = 2.57 s

Explanation:

The period of a wave is equal to the time it takes for one wavelength to pass by a fixed point.

No of waves observed = 10.9

Time taken, t = 28 s

We need to find the period of the water waves. Number of waves per unit time is called frequency. Let it is f. So,

[tex]f=\dfrac{n}{t}\\\\f=\dfrac{10.9}{28}\\\\f=0.389\ Hz[/tex]

If T is period of the wave. It is equal to the reciprocal of frequency.

[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.389}\\\\T=2.57\ s[/tex]

So, the period of the water waves is 2.57 seconds.