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3 years ago

6

A woman lifts a barbell 2.0 m in 5.0 s. If she lifts it the same distance in 10 s, the work done by her is:______

1 answer:

Zina [86]3 years ago

3 0

Answer:

d

Explanation:

none

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As Aubrey watches this merry-go-round for a total of 2 minutes, she notices the black horse pass by 15 times. What is the period

marishachu [46]

2 min = 120 sec

120/15 = 8

The black horse represents 8 seconds.

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3 years ago

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What happens to light when it changes speed?

Brut [27]

Hey there!

When light changes speed, it REFRACTS.
Your answer is going to be option C.

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3 years ago

A roller coaster's velocity at the top of the hill is 10 m/s. Two seconds later it reaches the bottom of the hill with a velocit

svet-max [94.6K]

your answer is 8 m / sec²

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3 years ago

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

3 years ago

A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t

Oduvanchick [21]

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

$T = \frac{2u Sin\theta }{g}$

Final horizontal component of velocity, vx = ux = u Cos 30

Let vy is teh final vertical component of velocity.

Use first equation of motion

vy = uy - gT

$v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}$

$v_{y}=u Sin 30 - 2u Sin 30$

vy = - u Sin 30

The magnitude of final velocity is given by

$v = \sqrt{v_{x}^{2}+v_{y}^{2}}$

$v = \sqrt{\left (uCos 30 \right )^{2}+\left (uSin 30 \right )^{2}}$

v = u

Thus, the velocity is same as it just reaches the ground.

6 0

4 years ago