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Illusion [34]
3 years ago
14

Help?? This is for my Physics class

Physics
1 answer:
Fittoniya [83]3 years ago
7 0
For distance vs time graphs, the slope stands for speed, as speed is a scalar, along with distance.
For displacement (position) vs time graphs, the slope stands for velocity, as velocity is a vector, along with displacement.
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Answer:

Centripetal force is the force that keeps the yoyo going in a circle, if the string breaks, the yoyo would would fly off in a direction that is different to the point on the circle.

8 0
3 years ago
A runner completes the 300-meter dash in 38 seconds. What is the speed of the runner? Round your answer to the nearest tenth.The
Galina-37 [17]
The speed of the runner is 300 m /38 seconds. You can simplify this answer to be about 7.9 m/s
7 0
3 years ago
A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 1
attashe74 [19]

The average act on her during the deceleration is 4.47 meters per second.                                                                

<u>Explanation</u>:

<u>Given</u>:

youngster mass m = 50.0 kg

She steps off a 1.00 m high platform that is s = 1 meter

She comes to rest in the 10-meter second

<u>To Find</u>:

The average force and momentum

<u>Formulas</u>:

p = m * v

F * Δ t = Δ p

vf^2= vi^2+2as

<u>Solution</u>:

a = 9.8 m/s

vi = 0

vf^2= 0+2(9.8)(1)

vf^2 = 19.6

vf = 4.47 m/s .

Therefore the average force is 4.47 m/s.                                          

                                 

5 0
3 years ago
•• CP Two blocks connected by a light horizontal rope sit at rest on a horizontal, frictionless surface. Block AA has mass 15.0
Firdavs [7]

Answer:

(a) T= 38.4 N

(b) m= 26.67 kg

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Kinematics

d= v₀t+ (1/2)*a*t² (Formula 2)

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

v₀=0, d=18 m , t=5 s

We apply the formula 2 to calculate the accelerations of the blocks:

d= v₀t+ (1/2)*a*t²

18= 0+  (1/2)*a*(5)²

a= (2*18) / ( 25) = 1.44 m/s² to the right

We apply Newton's second law to the block A

∑Fx = m*ax

60-T = 15*1.44

60 - 15*1.44 = T

T = 38.4 N

We apply Newton's second law to the block B

∑Fx = m*ax

T = m*ax

38.4 = m*1.44

m= (38.4) / (1.44)

m = 26.67 kg

7 0
3 years ago
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