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3 years ago

When is a body said to be motion (movement )?

Physics

1 answer:

Stels [109]3 years ago

3 0

A body is said to be in motion when it's position changes continuously.

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Which changes do not involve forming or breaking chemical bonds? rusting and cutting burning and melting burning and digesting b

r-ruslan [8.4K]

Answer:

boiling and melting!

Explanation:

These do not make any changes chemically to the substance.

Hope this helps!

6 0

3 years ago

A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f

Snezhnost [94]

Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

$\mu mg = m\frac{v^2}{r}$

where

$\mu$ is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

$v=\sqrt{\mu gr}$

And by substituting the data of the problem, we find the speed at which the car begins to skid:

$v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s$

7 0

3 years ago

Read 2 more answers

A 500 500500- kg kgstart text, k, g, end text object is accelerating to the right at 10 cm / s 2 10 cm/s 2 10, start text, space

enyata [817]

Given Information:

Mass = m = 500 kg

Acceleration = a = 10 cm/s²

Required Information:

Magnitude of rightward net force = F = ?

Answer:

Magnitude of rightward net force = 50 N

Explanation:

From the Newton's second law of motion

F = ma

Where m is the mass and a is the acceleration

To get force in Newtons first convert 10 cm/s² into m/s²

10/100 = 0.1 m/s²

F = 500*0.1

F = 50 N

Therefore, the magnitude of rightward net force acting on it is 50 Newtons.

8 0

3 years ago

Read 2 more answers

Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2

Alona [7]

Answer:

The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Explanation:

Worked : work can be defined as the product of force and distance.

The S.I unit of work is Joules (J).

Mathematically it can be represented as,

W = F×d.................. Equation 1

d = W/F.............................. Equation 2

where W = work, F = force, d = distance.

<em>Given: W = 12 J</em>

(i) for the 3.0 N weight,

using equation 2

d = 12/3

d= 4 m.

(ii) for the 4.0 N weight,

d = 12/4

d = 3 m.

(iii) for the 6.0 N weight,

d = 12/6

d = 2 m.

(iv) for the 2.0 N weight,

d = 12/2

d = 6 m

Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

8 0

3 years ago

Why is the moon's umbra much smaller during a solar eclipse?

Stolb23 [73]

Answer:

the sun is larger than the moon so it creates a shadow

7 0

2 years ago