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3 years ago

Figure 1 shows the motion of three balls. The curved paths followed by balls B and Care examples of

Physics

1 answer:

nlexa [21]3 years ago

7 0

Answer:

in brainly app there is an option that you can upload graph, plot or diagram related to your question. you can use that app to show diagram related to your question so that one can answer your question in better way

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2. What is the power rating of an engine capable of lifting a 100 kg object 5 m vertically

Leya [2.2K]

Work=f.d
Work=100*50 = 500
Power = work/time = 500/4
=125 watt

7 0

3 years ago

What phenomenon causes colors of visible light to be separated by a prism?

Allushta [10]

Dispersion im pretty sure

6 0

3 years ago

The spring in the muzzle of a child's spring gun has a spring constant of 730 N/m. To shoot a ball from the gun, first the sprin

Korolek [52]

Answer:

a. V=11.84 m/s

b.x=0.052m

Explanation:

a).

Given

$K=730 N/m$ , $m=0.053kg$ , $h=1.90m$ .

$v_f^2=v_i^2+2*g*h$

$v_i^2=2*g*h=2*9.8m/s^2*1.9m$

$v_i=\sqrt{2*9.8m/s^2*1.9m}=\sqrt{37.24 m^2/s^2}$

$v_i=6.1 m/s$

$v_i=V*sin(31)$

$V=\frac{v_i}{sin(31)}=\frac{6.1m/s}{sin(31)}$

$V=11.84 m/s$

b).

$K_k=\frac{1}{2}*K*x^2$

No friction on the ball so:

$x^2=\frac{2*K_k}{K}$

$x=\sqrt{\frac{2*0.053kg*9.8m/s^2*1.9m}{730N/m}}$

$x=\sqrt{2.7x10^{-3}m^2}=0.052m$

5 0

3 years ago

PLS HELP. A flatbed car of a train moves 8 m/s to the east. A jogger runs along to top the flatbed car (which is not very safe)

jonny [76]

Answer:

2 m/s

Explanation: Given that a flatbed car of a train moves 8 m/s to the east. A jogger runs along to top the flatbed car (which is not very safe) with a velocity of 6 m/s to the west.

Since the jogger is moving in an opposite direction to the direction of the train, and velocity is the distance covered in a specific direction, the jogger will be moving at a velocity relative to the velocity of the train.

Velocity = (8 - 6) m/s

Velocity = 2 m/s

Therefore, the jogger will be moving at the speed of 2 m/s

5 0

3 years ago

a 30 kg child is sitting 2 meters from the center of a merry go round. The coefficients of static and kinetic friction between t

laiz [17]

Answer: A

Explanation:

From the question, the given parameters are given.

Mass M = 30kg

Radius r = 2 m

Coefficient of static friction μ = 0.8

Coefficient of kinetic friction μ = 0.6

Kinetic friction Fk = μ × mg

Fk = 0.6 × 30 × 9.8

Fk = 176.4 N

The force acting on the merry go round is a centripetal force F.

F = MV^2/r

This force must be greater than or equal to the kinetic friction Fk. That is,

F = Fk

F = 176.4

Substitute F , M and r into the centripetal force formula above

176.4 = (30×V^2)/2

Cross multiply

352.8 = 30V^2

V^2 = 352.8/30

V = sqrt (11.76) m/s

V = 5.24 m/s

Therefore, the maximum speed of the merry go round before the child begins to slip is sqrt (12) m/s approximately

4 0

3 years ago