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4 years ago

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A paint brush is dropped from the top of a tall ladder and free falls to the ground. What changes, if any, would be observed of

the velocity and the acceleration of the brush as it falls? Pick the two answers. The velocity increases. The velocity decreases. The velocity remains a constant value. The acceleration increases. The acceleration decreases. The acceleration remains a constant value.

1 answer:

muminat4 years ago

4 0

Answer:

Explanation:

As the paint brush is dropped from the top of a tall ladder it started free falling

i.e. velocity of brush increases as it falls down due to the acceleration provided by gravity.

We know acceleration due to gravity is the attraction experienced by the object due to earth attraction

Value of acceleration due to gravity is constant i.e. $9.8\ m/s^2$

Therefore the correct choice is

velocity is increasing and acceleration is constant

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You measure an electric field of 1.36×106 N/C at a distance of 0.158 m from a point charge. There is no other source of electric

marysya [2.9K]

Answer:

The Electric flux will be $0.42\times10^6\ \rm N.m^2/C$

Explanation:

Given

Strength of the Electric Field at a distance of 0.158 m from the point charge is

$E=1.36\times10^6\ \rm N/C$

We know that the flux of the Electric Field can be calculated by using Gauss Law which is given by

$\int E.dA=\dfrac{q_{in}}{\epsilon_0}\\$

Let consider a sphere of radius 0.158 m as Gaussian Surface at a distance of 0.158 m from the point charge and Let $\phi$ be the flux of the Electric Field coming out\passing through it which is given by

$\phi=\int E.dA=1.36\times10^6 \times4\pi \times 0.158^2\\\\=0.42\times10^6\ \rm N.m^2/C$

It can be observed that same amount of flux which is passing through the Gaussian sphere of radius 0.158 is also passing through the Gaussian sphere of radius 0.142 m at a distance of 0.142 m from its centre.

Also it can be observed that the charge inside the two Gaussian Sphere mentioned have same value so the Flux of electric field through them will also be same.

So the electric flux through the surface of sphere that has given charge at its centre and that has radius 0.142 m is $0.42\times10^6\ \rm N.m^2/C$

8 0

3 years ago

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the f

Andreas93 [3]

Answer:

Part a)

$\theta_2 = 15 degree$

Part b)

$\Delta t = 2.88 s$

Explanation:

Part a)

In order to have same range for same initial speed we can say

$R_1 = R_2$

$\frac{v^2 sin2\theta_1}{g} = \frac{v^2 sin2\theta_2}{g}$

so after comparing above we will have

$\theta_1 = 90 - \theta$

so we have

$75 = 90 - \theta_2$

$\theta_2 = 15 degree$

Part b)

Time of flight for the first ball is given as

$T_1 = \frac{2vsin\theta}{g}$

$T_1 = \frac{2(20)sin75}{9.81}$

$T_1 = 3.94 s$

Now for other angle of projection time is given as

$T_2 = \frac{2(20)sin15}{9.81}$

$T_2 = 1.05 s$

So here the time lag between two is given as

$\Delta t = T_1 - T_2$

$\Delta t = 3.94 - 1.05$

$\Delta t = 2.88 s$

5 0

3 years ago

A car with a mass of 800 kg, moving at +15 m/s, crashes head on with a 1500 kg truck moving at -45.5 m/s in the opposite directi

Sloan [31]

Answer:

v = -24.5 m/s

Explanation:

Use conservation of momentum P = m*v. Unless there are external forces involved, total momentum must be conserved.

P: momentum

m: mass

v: velocity

1. calculate the total momentum before the collision:

$P=m_{car}v_{car} + m_{truck}v_{truck}$

2. the total momentum after the collision must not change:

$P_{before}=P_{after}$

3. after the collision the velocity of the car is the same as the velocity of the truck:

$v = v_{car} = v_{truck}$

4. combining 2. and 3.:

$P= (m_{car}+m_{truck})v$

5. this equation only has one unknown and can be solved for v:

$v = \frac{P}{m_{car}+m_{truck}} =\frac{m_{car}v_{car} + m_{truck}v_{truck}}{m_{car}+m_{truck}}$

$v=\frac{800*15 + 1500 * (-45.5)}{800 + 1500} = -24.5$

3 0

4 years ago

WHICH ONE!!! ASAP FOR A RETAKE FOR SCIENCE PLS

aleksandr82 [10.1K]

I’m 95% sure it’s covalent bonds.

8 0

3 years ago

Your friend turns the volume up on their speaker system, so that the sound intensity is 4 times greater than it was beforehand.

erica [24]

Answer:

4

Explanation:

We know that intensity I = P/A where P = power and A = area through which the power passes through.

Now, let the initial intensity of the speaker be I₀ and its initial power be P₀. Since the intensity is increased by a factor of 4, the new intensity be I and new power be P.

So, I = P/A and I₀ = P₀/A

Now, if I = 4I₀,

P/A = 4P₀/A

P = 4P₀

Now, energy E = Pt, where t = time. So, P = E/t and P₀ = E₀/t

Substituting P and P₀ into the equation, we have

P = 4P₀

E/t = 4E₀/t

E = 4E₀

Since the energy is four times the initial energy, the energy output increases by a factor of 4.

4 0

3 years ago