If you live on the path of a jet stream, will the temperature get hotter or colder?

How far does a 1.2g bullet with kinetic energy of 1.2 j go in 2 seconds ?<br><br><br>

almond37 [142]

Answer:

To calculate the energy in joules, simply enter the mass of ammunition (in grams) that you use, and the fps that you've read from your Chrono unit.

5 0

2 years ago

When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do

Semenov [28]

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force

$\Delta p$ is the change in momentum

$\Delta t=0.030 s$ is the time interval

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

$v=281 km/h =78.1 m/s$ is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N$

7 0

3 years ago

A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i

Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0

3 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago

A wheel is rotating freely at angular speed 420 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi

Mashcka [7]

Answer:60 rev/min

Explanation:

Given

angular speed of first shaft $\omega _1=420\ rev/min$

Moment of inertia of second shaft is seven times times the rotational speed of the first i.e. If I is the moment of inertia of first wheel so moment of inertia of second is 7 I

As there is no external torque therefore angular momentum is conserved

$L_1=L_2$

$I_1\omega _1=I_2\omega _2$

$I\times (420)=7 I\times (\omega _2)$

$\omega _2=\frac{420}{7}$

$\omega _2=60\ rev/min$

8 0

3 years ago