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ryzh [129]
3 years ago
12

Assign a positive velocity to the red box and negative velocity to the blue box. Are they moving in opposite direction?

Physics
1 answer:
attashe74 [19]3 years ago
4 0

Answer:

<em>Yes, they are moving in opposite direction one to the other.</em>

Explanation:

Velocity is a vector quantity, which means that it has both magnitude and direction. The magnitude shows the size of the velocity, and the direction shows which way it is moving in reference to a chosen reference direction. If the red box is assigned a positive velocity, and the blue box is assigned a negative velocity, as indicated in the question, then it means that the red box, and the blue box, both move in opposite direction to the other.

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A Ferris wheel is a vertical, circular amusement ride with radius 6.0 m. Riders sit on seats that swivel to remain horizontal. T
Oksana_A [137]

Answer:

Incomplete question

The complete question is

A Ferris wheel is a vertical, circular amusement ride with radius 6.0 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 9.6 s. Consider a rider whose mass is 96 kg.

At the bottom of the ride, what is the rate of change of the rider's momentum?

Explanation:

Radius of wheel is 6m

Rider mass=96kg

He completes one revolution in 9.6s

Let get angular velocity (w)

1 Revolution =2πrad

θ=2πrad

w= θ/t

w=2π/9.6

w=0.654rad/s

Linear speed is give as

v=wr

v=0.654×6

v=3.93m/s

Centripetal acceleration a

a=rw²

a=6×0.654²

a=2.57m/s²

Acceleration due to gravity g=9.81m/s²

According to Newton's second law of motion net force acting on the rider at the bottom of the ride is given by: the two force acting at the bottom is the normal and the weight of the rider

ΣF = ma

N-W=ma

N-mg=ma

N=ma+mg

N=m(a+g)

N=96(2.57+9.81)

N=1188.48 N

Therefore the rate of change of momentum at the bottom of the ride is 1188.48 N.

8 0
3 years ago
A race car travels 44.3 m/s around a banked (45° with the horizontal) circular (radius = 200 m) track. What is the magnitude of
djverab [1.8K]

The magnitude of the resultant force is given by the centripetal force, since the car is under a circular motion. So, we have:

F_c=ma_c

The centripetal acceleration is given by:

a_c=\frac{v^2}{r}

Where v is the linear speed and r the radius of the circular motion. Replacing this and solving:

F=m\frac{v^2}{r}\\F=80kg\frac{(44.3\frac{m}{s})^2}{200m}\\F=785N*\frac{1kN}{1000N}\\F=0.785kN

3 0
3 years ago
1.How does gravity change your weight?<br><br> 2.How do force work in nature?
kodGreya [7K]

In a stronger gravitational field a given mass will have a larger weight.

3 0
3 years ago
A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a
IrinaK [193]

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>
  • Range = U²× sin(2θ)/g
  • U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
  • U=√{Range×g/sin(2θ)}
  • Here, range= 2.20m, = 36.5°
  • U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>
  • Time of flight= (2×U×sinθ)/g
  • So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

#SPJ1

4 0
2 years ago
A 10 m long uniform beam weighing 100 N is supported by two ropes at the ends. If a 400 N person sits at 2.0 m from one end of t
Naddik [55]

Answer:

T1 = 130N, T2 = 370N

Explanation:

In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.

1. forces:

Let tension in rope 1 be T1 and in rope 2 be T2:

ma = T1 + T2 - 100N - 400N = 0

(1) T1 + T2 = 500N

2. torque around the center point of the beam:

τ = r x F = 5*T1 + 3*400N - 5*T2 = 0

(2) T1 - T2 = -240N

Solving both equations:

T1 = 130N

T2 = 370N

3 0
3 years ago
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