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3 years ago

A 30 kg child is in an elevator accelerating down at a rate of -5 m/s2.

Physics

1 answer:

Annette [7]3 years ago

6 0

Answer:

The force is - 150 N

Explanation:

F = 30 • - 5 = - 150

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If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c

Cloud [144]

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction $\mu =0.8$

Angle of inclination is equal to $\Theta =tan^{-1}\mu$

$\Theta =tan^{-1}0.8=38.65^{\circ}$

$tan{38.65^{\circ}}=0.8$

Radius is given r = 28 m

Acceleration due to gravity $g=9.8m/sec^2$

We know that $tan\Theta =\frac{v^2}{rg}$

$0.8=\frac{v^2}{28\times 9.8}$

$v^2=219.52$

$v=14.816m/sec$

So before start of slide velocity will be 14.81 m/sec

3 0

3 years ago

A person walks 4.0 kilometers north, then 4.0 kilometers west. His displacement is closest to

larisa86 [58]

Use the Pythagoras for the magnitude and the tan^-1 x = -1 for the angle

displacement = 4^2 + 4^2 = 32 = 4 sqrt(2) = 5.65 km

angle is 135 degrees.

7 0

3 years ago

3. A model rocket is launched straight upward at 58.8 m/s.

kolezko [41]

Answer:

a). 6 seconds

b). 12 seconds

c). 176.4 meters

Explanation:

a). Equation to be applied to calculate the time taken by the rocket to reach at the peak height,

v = u - gt

where v = final velocity

u = initial velocity = 58.8 m per sec

g = gravitational pull = 9.8 m per sec²

t = duration of the flight

At the peak height,

v = 0

Therefore, 0 = 58.8 - (9.8)(t)

t = $\frac{58.8}{9.8}$

= 6 seconds

b). Total time of flight = 2(Time taken to go up)

= 2×6

= 12 sec

c). Formula to get the peak height is,

$h=ut-\frac{1}{2}gt^2$

h = (58.8)6 - $\frac{1}{2}(9.8)(6)^2$

= 352.8 - 176.4

= 176.4 meters

8 0

4 years ago

Which of the following is the energy of motion?

Bogdan [553]

Kinetic energy is the energy of the motion.

This is because the kinetic energy can be calculated by using the formula,

K.E.=1/2 mv^2

where, v is the velocity of moving body i.e. the body is in motion.

8 0

4 years ago

Outside the space shuttle, you and a friend pull on two ropes to dock a satellite whose mass is 700 kg. The satellite is initial

MaRussiya [10]

Answer:

your friend did a work of 1717.34 J

Explanation:

The sum of the work made by your friend and you is equal the change in the kinetic energy, so:

$W_a+W_b=\frac{1}{2}mv_f^{2}-\frac{1}{2}mv_i^{2}$

Where $W_a$ is the work that you did, $W_b$ is the work that your friend did, m is the mass of the satellite and $v_f$ and $v_i$ are the final and initial speeds

Additionally the work that you did is equal to:

$W_a=F.s$

Where F and s are vectors. So, the vector s that said the displacement is calculated as:

s = <2.3, 2.6, -11.8> - <3.3, -1.6, 2.0>

s = <-1 , 4.2 , -13.8>

Therefore, $W_a$ is:

$W_a=.$

$W_a=(-400*-1)+(500*4.2)+(250*-13.8)\\W_a=400+2100-3450\\W_a=-950$

Then, replacing the values on initial equation and solving for $W_b$ , we get:

$-950+W_b=\frac{1}{2}(700)(6.18)^{2}-\frac{1}{2}(700)(6)^{2}\\-950+W_b=767.24\\W_b=767.24+950\\W_b=1717.34J$

Finally, your friend did a work of 1717.34 J

3 0

3 years ago