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3 years ago

7

How does the range change as the angle increases?

1 answer:

RSB [31]3 years ago

5 0

It first increases then it decreases give me brainlest please

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Mary and her younger brother Alex decide to ride the 26 ft diameter carousel at the State Fair. Mary sits on one of the horses i

hammer [34]

Answer:

$a_M=1.92a_A$

Explanation:

$\omega_M=\omega_A$ = Angular speed

$r_M$ = Distance of Mary = 11.5 ft

$r_A$ = Distance of Alex = 6 ft

Ratio of centripetal acceleration is given by

$\dfrac{a_M}{a_A}=\dfrac{\omega_M^2r_M}{\omega_A^2r_A}\\\Rightarrow \dfrac{a_M}{a_A}=\dfrac{r_M}{r_A}\\\Rightarrow a_M=a_A\dfrac{r_M}{r_A}\\\Rightarrow a_M=\dfrac{11.5}{6}a_A\\\Rightarrow a_M=1.92a_A$

Mary's centripetal acceleration is 1.92 times the centripetal acceleration of Alex

8 0

3 years ago

A track star runs the first 500 meters of a race in only 75 seconds. What is the track

irina1246 [14]

Answer:

6.667 m/s

Explanation:

500 meters = 75 seconds

500 meters / 75 seconds = 6.667 m/s

7 0

3 years ago

Read 2 more answers

maxonik [38]

D. Was a leader in the woman's suffrage movement

7 0

3 years ago

When heat flows from one substance to another, what happens to the temperature of the substance giving off the heat and to the t

blagie [28]

The temperature of the substance giving off the heat decreases while the temperature of the substance receilving the heat increases. they leach what is called equlibrium point where heat energy can longer be exchanged hence equql temperature. this isThermal physics

8 0

3 years ago

The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0

3 years ago