Two men, Joel and Jerry, push against a car that has stalled, trying unsuccessfully to get it moving. Jerry stops after 10 min, while Joel is able to push for 5.0 min longer. Compare the work they do on the car.

a. Joel does 75% more work than Jerry.
b. Joel does 25% more work than Jerry.
c. Jerry does 50% more work than Joel.
d. Joel does 50% more work than Jerry
e. None of the above .

Answer:

The answer is "0.5555 m"

Explanation:

Where the reference leaves the list and the viewer is at rest:

[tex]\lambda'=\frac{v-v_s}{v} \times \lambda\\\\[/tex]

   [tex]=\frac{343 \frac{m}{s} - (-62.4 \frac{m}{s})}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{343 \frac{m}{s} + 62.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m[/tex]

   [tex]=0.5555 \ m[/tex]

Answer:

0.159

Explanation:

Given that

F1 = 440 N

F2 = 340 N

m = 500 kg

acceleration, a = 0 m/s²

Remember that F = m.a

If F - f(f) = 0, then F = f(f)

N = mg

f(f) = μN, substituting for N, we have

f(f) = μmg

The total force, F = F1 + F2

F = 440 + 340

F = 780 N

Recall that we'd already proven that

F = f(f), so F = 780 N = f(f)

And again, f(f) = μmg, if we substitute for all the values, we have

780 = μ * 500 * 9.81

780 = μ * 4905

μ = 780 / 4905

μ = 0.159

Therefore, the coefficient of static friction of the crate on the floor is 0.159

Answer:

Approximately [tex]3.99\times 10^{4}\; \rm J[/tex] (assuming that the melting point of ice is [tex]0\; \rm ^\circ C[/tex].)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

[tex]\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}[/tex]

The energy required comes in three parts:

The following equation gives the amount of energy [tex]Q[/tex] required to raise the temperature of a sample of mass [tex]m[/tex] and specific heat capacity [tex]c[/tex] by [tex]\Delta T[/tex]:

[tex]Q = c \cdot m \cdot \Delta T[/tex],

where

For the first part of energy input, [tex]c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:

[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].

Calculate the energy required to achieve that temperature change:

[tex]\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}[/tex].

Similarly, for the third part of energy input, [tex]c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:

[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].

Calculate the energy required to achieve that temperature change:

[tex]\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}[/tex].

The second part of energy input requires a different equation. The energy [tex]Q[/tex] required to melt a sample of mass [tex]m[/tex] and latent heat of fusion [tex]L_\text{f}[/tex] is:

[tex]Q = m \cdot L_\text{f}[/tex].

Apply this equation to find the size of the second part of energy input:

[tex]\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}[/tex].

Find the sum of these three parts of energy:

[tex]\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}[/tex].

Answer:

By using [tex]v^{2}_{f} = v^{2}_{i} + 2a\Delta y[/tex], with [tex]v_{i} = -12m/s[/tex] and [tex]\Delta y = -40m[/tex]:

[tex]v^{2}_{f} = v^{2}_{i} + 2a/Delta y[/tex]

[tex]v^{2} = (-12m/s)^{2} + 2(-9.80m/s^{2})(-40m)[/tex]

[tex]v = -30m/s[/tex]

Explanation:

Hope this helped!

A rock is thrown downward from the top of a 40 m tall tower, with The speed of the rock just before hitting the ground will be equal to -30 m/s.

Friction is the resistance to a thing moving or rolling over another solid object. Although frictional forces can be advantageous, such as the traction required to walk while slipping, they also provide a significant amount of resistance to motion. In order to overcome frictional resistance in the moving parts, about 20% of an automobile's engine power is used.

The forces of attraction, also referred to as adhesion, between the contact zones of the surface, which are always minutely uneven, seem to be the main contributor to friction between metals.

From the given information in the question,

v²(f) = v²(i) + 2aΔy

v² = (-12 m/s)² + 2(-9.8)(-40)

v = -30 m/s.

Therefore, the velocity of the rock is -30 m/s.

To know more about Friction:

https://brainly.com/question/13000653

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Answer: The height of the cliff is 104.59 m

Explanation:

The horizontal speed of the car when it leaves the cliff is 13 m/s, and it hits the ground 60m from the shoreline.

Here we can use the relationship:

Time*Speed = Distance.

To find the time that the car is in the air, we know that:

speed = 13m/s

distance = 60m

time = T

13m/s*T = 60m

T = (60m)/13m/s = 4.62 s

This means that the car is falling for 4.62 seconds.

Now let's analyze the vertical problem.

As the car leaves the cliff, it only has horizontal velocity, this means that the vertical initial velocity will be zero

The only force acting in the vertical axis is the gravitational force, this means that the acceleration will be equal to the gravitational acceleration, which is:

g = 9.8m/s^2

then:

a = -9.8m/s^2

Where the negative sign is because the acceleration is pulling the car downwards.

To get the vertical velocity, we could integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the constant of integration and the initial vertical velocity, that we already know that is equal to zero, then the vertical velocity as a function of time can be written as:

v(t) = (-9.8m/s^2)*t

To get the vertical position equation, we need to integrate again over the time:

P(t) = (1/2)*(-9.8m/s^2)*t^2 + H

Where H is the constant of integration and the initial vertical position, then H will be the height of the cliff.

We know that the car needs 4.62 seconds to hit the ground, this means that:

P(4.6s) = 0m

Then:

P(t) = (1/2)*(-9.8m/s^2)*(4.62s)^2 + H = 0

            (-4.9m/s^2)*(4.62s)^2 + H = 0

          H =  (4.9m/s^2)*(4.62s)^2 = 104.59 m

This means that the cliff is 104.59 meters high

Answer:

The second one I think

Explanation:

B

Answer:

zoom in

Explanation:

Answer:

2.51 kg * m/s

Explanation:

In order to find momentum, use the equation below:

momentum = mass * velocity.

Since neither mass nor velocity was given, you must solve for both variables.

In order to solve for mass, use the force equation for its weight / gravitational force.

Fg (gravitational force) = 3.2 N = ma = 9.8m

mass = 3.2 N / 9.8 m/s^2 = 0.326531 kg

In order to solve for velocity, use the equation:

velocity = displacement / time

velocity = 10m / 1.3 s = 7.69231 m/s

Momentum = mass * velocity = 0.326531 kg *  7.69231 m/s = 2.51177 kg * m/s = 2.51 kg * m/s

Answer:

Dark matter:

- Doesn't interact with baryonic matter.

- It has not been observed directly

- dark matter does not absorb, reflect or even emit light, thereby making it to be extremely hard to spot. Therefore, it does not interact with electromagnetic radiation.

Explanation:

Dark matter:

- Doesn't interact with baryonic matter.

- It has not been observed directly

- dark matter does not absorb, reflect or even emit light, thereby making it to be extremely hard to spot. Therefore, it does not interact with electromagnetic radiation.

Baryonic matter:

- Has been observed directly because it includes nearly all the matter that we see in the world daily.

- It interacts with baryonic matter

- interacts with electromagnetic radiation

Dark Matter:

Baryonic Matter:

Dark matter can be defined as often invisible substances that are difficult to spot because they don't absorb, emit or reflect light.

Hence, dark matter do not affect human view because they do not interact or interfere with electromagnetic radiation (force).

Although, humans can see right through the (weakly interacting) dark matter but it has not been observed directly.

Baryonic matter can be defined as a dark matter that is made up of baryons such as neutrons, and protons. Also, they are ordinary matter (both fermions and hadrons), as distinct from exotic forms.

In conclusion. baryonic matter has been observed directly and it can interact with electromagnetic radiation.

Read more: https://brainly.com/question/15677526

Answer:

0.8 m/s

Explanation:

F(avg) * Δt = I, where

F(avg) = Average force

Δt = change in time

I = impulse

From the question, we know that the

average force is given as -17600 N

time change is given as 55 milliseconds

speed of the player, v = 8 m/s

Mass of the player is given as 110 kg

Impulse, I = F(avg) * Δt

I = -17600 * 0.055

I = -968

Now, using the Impulse-Momentum theory, we have that

Δp = I and thus,

Δp = m [v(f) - v(i)], where

Δp = change in the momentum

v(f) = final speed of the player

v(i) = initial speed of the player

Substituting the values, we have

I = m [v(f) - v(i)]

-968 = m.v(f) - m.v(i)

m.v(f) = m.v(i) - 968

110v(f) = 110 * 8 - 968

110v(f) = 880 - 968

110v(f) = -88

v(f) = -88 / 110

v(f) = -0.8 m/s

Answer:

angle of reflection and angle of incident is always equal

Answer:

The frequency stays the same and the wavelength decreases.

Explanation:

When sound wave enters a new medium where sound travels faster, its frequency will remain same because it depends only on the source.

The relation between wavelength and speed is inverse, it means when the speed of sound increases, its wavelength will decrease.

So, the frequency stays the same and the wavelength decreases. Hence, the correct option is (d).

They could determine it by counting their total amount of miles mark went by the directions he went to walk those miles.

Answer:

yo can someone answer i need this

Explanation:

Answer:

The value is [tex]E = 1.35 *10^{14} \ J[/tex]

Explanation:

From the question we are told that

    The mass of matter converted to energy on first test is  [tex]m = 1 \ g = 0.001 \ kg[/tex]

    The mass of matter converted to energy on second test [tex]m_1 = 1.5 \ g = 1.5 *10^{-3} \ kg[/tex]

    Generally the amount of energy that was released by  the explosion is  mathematically  represented as  

         [tex]E = m * c^2[/tex]

=>       [tex]E = 1.5 *10^{-3} * [ 3.0 *10^{8}]^2[/tex]

=>       [tex]E = 1.35 *10^{14} \ J[/tex]

Answer:

See the answers and the explanation below.

Explanation:

To solve these questions we must understand that speed is the relationship between the displacement and the time that the displacement lasts.

1. 300 [km], in 5 [hr]

d = displacement [km]

[tex]v = d/t[/tex]

where:

v = velocity [km/h]

t = time [hr]

[tex]v = 300/5\\v = 60 [km/h][/tex]

In one hour : [tex]d = 60*1\\d = 60 [km][/tex]

b. Car B 100 [km] in 2 [hr]

[tex]v =100/2\\v = 50 [km/h]\\[/tex]

C. The car A has the greatest average speed.

Answer:

true

Explanation:

Answer:

Since binary is only 1 and 0, you can use a flashlight to display something similar to Morse code (see explanation below)

Explanation:

In binary, 1 means "on" and 0 means "off". A way you can use visible light is through turning on and off a flashlight. If the flashlight is turned on, it would represent a 1. If the flashlight is turned off, it would represent a 0. To make the message easier and more accurately understood for the receiver make sure to flash the lights in a consistent pattern (ex. each flash lasts no longer than half a second, one second between each digit, etc.)

For example, let's say you're trying to send the message "11001"

  on     on    off     off     on

0       1       2       3       4       5      Numbers represent seconds

As you can see above the message starts at 0 seconds. Between 0 and 1 seconds the flashlight is turned on once. Between 1 and 2 seconds the flashlight is turned on again, Between 2 and 3 seconds as well as 3 and 4 seconds the flashlight is not turned on at all. And finally between 4 and 5 seconds the flashlight is turned on.

Answer:

(A)

Explanation:

Answer:

275 x 10"N

Explanation: