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3 years ago

What does the thermosphere and stratmosphere have in common

1 answer:

5 0

these include the troposphere (0 to 16 km), stratosphere (16 to 50 km), mesosphere (50 to 80km) and thermosphere (80 to 640km). The boundaries between these four layers are defined by abrupt changes in temperature, and include respectively the tropopause, stratopause and mesopause.

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A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the

Liula [17]

Answer:

The average force on ball by the golf club is 340 N.

Explanation:

Given that,

Mass of the golf ball, m = 0.03 kg

Initial speed of the ball, u = 0

Final speed of the ball, v = 34 m/s

Time of contact, $\Delta t=0.003\ s$

We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

$F=\dfrac{\Delta p}{\Delta t}\\\\F=\dfrac{mv-mu}{\Delta t}\\\\F=\dfrac{mv}{\Delta t}\\\\F=\dfrac{0.03\ kg\times 34\ m/s}{0.003\ s}\\\\F=340\ N$

So, the average force on ball by the golf club is 340 N.

4 0

4 years ago

An automobile starter motor has an equivalent resistance of 0.0510 Ω and is supplied by a 12.0 V battery with a 0.0090 Ω interna

Alisiya [41]

Answer:

a) 200A

b) 10.2V

c) 2.04kW

I=80A

V=4.08V

P=0.326kW

Explanation:

Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:

$I=\frac{V}{R}$

Where R is the equivalent resistance of the resistors in series

$R=0.0510+0.0090=0.0600[ohm]$

$I=\frac{12.0}{0.0600}=200A$

To calculate the voltage dropped by the motor we have to apply the voltage divider rule:

$V_m=V*\frac{R_m}{R_m+R_s}\\V_m=12.0*\frac{0.0510}{0.0600}\\V_m=10.2V$

The power dissipated supplied to the motor is given by:

$P=I^2*R_m\\P=(200)^2*0.0510=2.04kW$

now solving adding a 0.0900 ohm resistor:

$I=\frac{12.0}{0.15}=80A$

$V_m=12.0*\frac{0.0510}{0.15}\\V_m=4.08V$

$P=I^2*R_m\\P=(200)^2*0.0510=0.326kW$

5 0

3 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,