What acceleration will a 1000-kg car experience if a force of 500 N is applied?

A Russian athlete lifts 300kg to a height of 2.5 m in 2 seconds. What is the work done? And what power is exerted?

snow_tiger [21]

We want to find the work done and power exerted, let’s start with work first.

We know that the equation for work is: W = F * D. We need to find the force which we can find by using: F = M * A.

Mass: 300kg
Acceleration (using equation from photo): 1.25 m/s^2

(The equation says x but can be used with y values)

If you are confused about how I found the acceleration; I plugged in 2.5 for the final y value, 0 for the initial y value, 0 for the initial velocity and 4 for t squared.

To solve, for acceleration it’s a matter of simple algebra. You can subtract the initial y position and the initial velocity from the final y position because they are 0. This leaves you with 2.5 m = 1/2a * t^2, from here I multiplied 2.5 by 2 to get rid of the 1/2. Now I have 5 = a * t^2. T^2 is just 2 squared, so four. Simply divide 5 by 4, and boom, you get 1.25 m/s^2.

Force = 300 kg * 1.25 m/s^2 = 375 Newtons

So, work = 500 N * 2.5 m = 1000 Joules

Power: W/t

So, Power = 1000 J / 2 seconds = 500 Watts

Hope this helps!

5 0

3 years ago

a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?

evablogger [386]

<h2>Hello!</h2>

The answer is: 8.62m

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

$E=PE+KE$

Where:

$PE=m.g.h$

$KE=\frac{1*m*v^{2}}{2}$

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

$KE_{1}+PE_{1}=KE_{2}+PE_{2}$

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

$\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}$

So,

$h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}$

Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

5 0

4 years ago

Although the use of absorbances at 450 nm provided you with maximum sensitivity, the absorbances at, say, 400 nm or 500 nm are n

AleksAgata [21]

Answer:

Yes, the value will be the same.

Explanation:

Yes, or at least to some degree, that value of K will remain the same. You're looking for a difference in absorbance, and the difference should be visible at all wavelengths, not only at the limit. That being said, resolution varies, and if we don't read the value to the maximum, we can get a less accurate reading.

4 0

3 years ago

A machine is supplied energy at a rate of 4,000 W and does useful work at a rate of 3,760 W. What is the efficiency of the machi

Dmitry_Shevchenko [17]

= (3,760 joule/sec) / (4,000 joule/sec)

= 3,760 / 4,000 = 0.94 = 94%

8 0

4 years ago

find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4

Dima020 [189]

The rate of energy radiated by the man is 3.86 x $10^{-8}$ J/s. $m^{2}$ .

The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:

Q(t) = Aeσ $T^{4}$

where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.

To determine the rate of energy radiated by the man in the given question;

$\frac{Q(t)}{T^{4} }$ = Aeσ

But A = 1.7 m², e = 0.4 and σ = 5.67 x $10^{-8}$ J/s.

So that;

$\frac{Q(t)}{T^{4} }$ = 1.7 * 0.4 * 5.67 x $10^{-8}$

= 3.8556 x $10^{-8}$

= 3.86 x $10^{-8}$ J/s. $m^{2}$

Thus, the rate of energy radiated by the man is 3.86 x $10^{-8}$ J/s. $m^{2}$ .

Learn more on energy radiation of objects by visiting: brainly.com/question/12550129

4 0

3 years ago