Ask question

Ask question

All categories

3 years ago

13

Olympic gold medalist Michael Johnson runs one time around the track 400 meters in 38 seconds what is his displacement what is h

is average velocity

1 answer:

Sveta_85 [38]3 years ago

8 0

Displacement = 0, assuming that he runs back to original position
Average velocity is displacement/ time, since displacement =0, velocity is also 0

You might be interested in

A transformer has 100 turns in its primary coil and 75 turns in its secondary coil. If the input voltage is 12.0 V, what is the

Sindrei [870]

Is 54 por que es igual

7 0

3 years ago

A hoop is rolling without slipping along a horizontal surface with a forward speed of 5.50 m/s when it starts up a ramp that mak

Burka [1]

Answer:

$v_f = 4.22 m/s$

Explanation:

As we know by energy conservation

initial total energy = final total energy

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv_f^2 + \frac{1}{2}I\omega_f^2 + mgL sin\theta$

so we have

$v = R \omega$

$I = mR^2$

$mv_i^2 = mv_f^2 + mgL sin\theta$

$5.50^2 = v_f^2 + (9.81)(3) sin25$

$30.25 - 12.43 = v_f^2$

$v_f = 4.22 m/s$

5 0

4 years ago

In a skating stunt known as "crack-the-whip," a number of skaters hold hands and form a straight line. They try to skate so that

4vir4ik [10]

Answer:

The magnitude of the centripetal force that acts on him

Explanation:

Given that,

Mass = 80.0 kg

Distance = 6.10 m

Speed = 6.80 m/s

We need to calculate the magnitude of the centripetal force that acts on him

Using formula of the centripetal force

$F_{c}=\dfrac{mv^2}{r}$

Where, F = force

m = mass

v = speed

r = distance

Put the value into the formula

$F_{c}=\dfrac{80.0\times(6.80)^2}{6.10}$

$F_{c}=606.4\ N$

Hence, The magnitude of the centripetal force that acts on him

7 0

3 years ago

25 POINTS!! ANSWER CORRECTLY FOR BRANLIEST!

sertanlavr [38]

Answer:

the answer is probably most likely D first and if thats incorrect than its B

♡ <em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>♡</em>

4 0

3 years ago

the system shown above is released from rest. if friction is negligible, the acceleration of the 4.0 kg block sliding on the tab

JulsSmile [24]

The acceleration of the first block (4 kg) is -9.8 m/s².

The given parameters:

<em>Mass of the first block, m₁ = 4.0 kg</em>
<em>Mass of the second block, m₂ = 2.0 kg</em>

The net force on the system of the two blocks is calculated as follows;

$m_2 g - T = m_1 a$

where;

<em>T </em><em>is the tension in the connecting string due weight of the first block</em>

$m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2$

Thus, the acceleration of the first block (4 kg) is -9.8 m/s².

Learn more about net force on two connected blocks here: brainly.com/question/13539944

5 0

3 years ago