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2 years ago

Suppose a car traveled 120 miles in two hours. What can we conclude about the car based on the mean value theorem.

Physics

1 answer:

Lapatulllka [165]2 years ago

4 0

We can conclude that it is traveling at a constant rate of 60 miles per hour
120mi/2hrs
= 60

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Light with a wavelength of 400 nm strikes the surface of cesium in a photocell, and the maximum kinetic energy of the electrons

Firdavs [7]

Answer:

The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

Explanation:

Given that,

Wavelength = 400 nm

Energy $E=1.54\times10^{-19}\ J$

We need to calculate the longest wavelength of light that is capable of ejecting electrons from that metal

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$\lambda=\dfrac{hc}{E}$

Put the value into the formula

$\lambda=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.54\times10^{-19}}$

$\lambda=1292\times10^{-9}\ m$

$\lambda=1292\ nm$

Hence, The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

8 0

3 years ago

Which person is weightless?

ahrayia [7]

Answer: Option (b) is the correct answer.

Explanation:

The force of gravity acting on an object helps in determining the weight of an object. But a place where there will be no gravity or have zero gravitational pull then it means the person will be weightless.

For example, force of gravity on moon is zero which means any object or person on moon will be weightless.

On the other hand, when a child is in the air as she plays on a trampoline then it means gravitational pull form the earth is acting on it. So, it will definitely has some weight.

Similarly, a scuba diver exploring a deep-sea wreck is under the ground where there will be force of gravity. Hence, it will also have some weight.

Thus, we can conclude that an astronaut on the Moon is the person who is weightless.

7 0

3 years ago

A lead nucleus is spherical with a radius of about 7 ✕ 10⁻¹⁵ m. The nucleus contains 82 protons (and typically 126 neutrons). Be

Komok [63]

Answer:

∈= $3.1584x10^{26} \frac{V}{m}$

Explanation:

Using the Gauss Law to determine the electric field of the net flux at the surface of the nucleus

∈ $=\frac{P}{E_{o}}$

The P is the charge density and 'Eo' is the constant of permittivity in free space

to find P

$P=\frac{q}{V}$

$V=\frac{4}{3}*\pi*r^3$

$V=\frac{4}{3}\pi*(7x10^{-15})^3$

$V=2.932x10^{-14} m^3$

$P=\frac{82C}{2.932x10^{-14}m^3}=2.7965x10^{15} \frac{C}{m^3}$

So replacing

∈ $=\frac{2.7965x10^{15}\frac{C}{m^3}}{8.8542x10^{-12}\frac{C^2}{N*m^2}}$

∈= $3.1584x10^{26} \frac{V}{m}$

3 0

3 years ago

The two forces in each pair can either both act on the same body or they can act on different bodies. The two forces in each pai

kotykmax [81]

Answer: The given statement is false.

Explanation:

According to Newton's third law of motion, every action has an equal and opposite reaction. So, when we apply force in one direction on an object then the object also applies a force in the opposite direction.

Hence, it is true that two forces in each pair of forces act in opposite directions.

For example, when we push a wooden box of 20 kg in the forward direction then the box will also apply a force in the opposite direction.

But the statement two forces in each pair can either both act on the same body or they can act on different bodies is false.

5 0

3 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

3 years ago