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mezya [45]
3 years ago
9

If a body with a mass of 4 kg is moved by a force of 20 N, what is the rate of its acceleration?​

Physics
1 answer:
valentina_108 [34]3 years ago
8 0

Explanation:

F = ma

20 N = (4 kg) a

a = 5 m/s²

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In handball, who is the only player allowed in the goalie area?
lidiya [134]

Answer:

Only the goalie is allowed inside the goal crease. The only exception when another player is allowed in the goal area is when they take off from outside the goal area, and shoots or passes the ball before landing. To avoid interference with other players, the player must then exit the goal area as soon as possible.

Explanation:

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3 years ago
a block measuring 20cm by 10cm by 5cm rests on a flat surface. The block has a weight of 3N. Determine the maximum pressure it e
Lady_Fox [76]
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4 0
4 years ago
If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can st
liberstina [14]

Answer:

The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Explanation:

Given;

coefficient of kinetic friction, μ = 0.84

speed of the automobile, u = 29.0 m/s

To determine the  the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration

x is the shortest distance

First we determine a;

From Newton's second law of motion

∑F = ma

F is the kinetic friction that opposes the motion of the car

-Fk = ma

but, -Fk = -μN

-μN = ma

-μmg = ma

-μg = a

- 0.8 x 9.8 = a

-7.84 m/s² = a

Now, substitute in the value of a in the equation above

v² = u² + 2ax

when the automobile stops, the final velocity, v = 0

0 = 29² + 2(-7.84)x

0 = 841 - 15.68x

15.68x = 841

x = 841 / 15.68

x = 53.64 m

Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

4 0
3 years ago
A 15.0 cm object is 12.0 cm from a convex mirror that has a focal length of -6.0 cm. What is the height of the image produced by
Vlad [161]

Answer: D. 5cm

Explanation:

Given the following :

Focal length (f) = - 6.0 cm

Height of object = 15.0cm

Distance of object from mirror (u) = 12.0cm

Height of image produced by the mirror =?

Firstly, we calculate the distance of the image from the mirror.

Using the mirror formula

1/f = 1/u + 1/v

1/v = 1/f - 1/u

1/v = 1/-6 - 1/12

1/v = - 1/6 - 1/12

1/v = (- 2 - 1) / 12

1/v = - 3 / 12

v = 12 / - 3

v = - 4

Using the relation :

(Image height / object height) = (- image distance / object distance)

Image height / 15 = - (-4) / 12

Image height / 15 = 4 / 12

Image height = (15 × 4) / 12

Image height = 60 / 12

Image height = 5cm

6 0
3 years ago
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