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2 years ago

Which statement correctly applied to a spinning mobile that changes its velocity over a period of time?

Physics

1 answer:

Yanka [14]2 years ago

8 0

Answer:

forces are acting on it

Explanation:

when a body is spinning and is changing its velocity per unit, the body is said to be acted upon by forces which causes the spinning.

a body won't change in velocity unless acted by forces.

for a spinning body, a centripetal and centrifugal forces are part of those forces responsible for the spinning car and causes a shift in its velocity.

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A 7.0-kilogram cart, A, and a 3.0-kilogram cart, B, are initially held together at rest on a horizontal, frictionless surface. W

7nadin3 [17]

For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively

(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s

<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>

7 0

2 years ago

A Roller Derby Exhibition recently came to town. They packed the gym for twoconsecutive weekend nights at South's field house. O

Alla [95]

Answer:

14.4 m/s

Explanation:

mass of Anna (Ma) = 68 kg

speed of Anna (Va) = 17 m/s

mass of SandraDay (Ms) = 76 kg

speed of SandraDay (Vs) = 12 m/s

We can find their speed (V) immediately after collision from the conservation of momentum where

(Ma x Va) + (Ms + Vs) = (Ma + Ms) x V

where V = speed immediately after collision

(68 x 17) + (76 + 12) = (68 + 76) x V

2068 = 144 V

V = 2068 / 144 = 14.4 m/s

8 0

2 years ago

Small blocks, each with mass m, are clamped at the ends and at the center of a rod of length L and negligible mass. Compute the

fenix001 [56]

Answer:

Answered

Explanation:

a) Two balls are at a distance of L/2 from the axis of rotation and one block at the center. ( center of rod).

therefore,

$I=2\times m\frac{L}{2}^2$

$I= \frac{1}{2}mL^2$

b) two balls at a distance L/4 at the from the axis and 1 ball at a distance 3L/4 from the from the axis.

$I= 2\times m\times(L/4)^2 + m(\frac{3L}{4})^2$

= $\frac{1}{16} mL^2(2+9)= \frac{11}{16}mL^2$

5 0

2 years ago

Compare and contrast the strength of the forces between two objects with a mass of 1 kg each, a charge of 10

DochEvi [55]

Answer:

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

Explanation:

Let's start by calculating each force.

Gravitational force

F = $G \frac{m_1m_2}{r^2}$

let's calculate

F = 6.67 10⁻¹¹ 1 1 / 1²

F = 6.67 10⁻¹¹ N

Electric force

F = $k \frac{q_1q_2}{r^2}$

indicates that the charge is q = 10 C

F = 9 10⁹ 10 10 / 1²

F = 9 10¹¹ N

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

3 0

2 years ago

I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =

Helen [10]

Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)

7 0

2 years ago