if a bowling ball hits a wall a force of 6 N, the wall exerts a force of how much back. on the bowling ball

To get a feeling for inertial forces discuss the familiar cases of accelerating in a car in a straight line while increasing or

inn [45]

Answer:

Explanation:

When we accelerate in a car on a straight path we tend to lean backward because our lower body part which is directly in contact with the seat of the car gets accelerated along with it but the upper the upper body experiences this force later on due to its own inertia. This force is accordance with Newton's second law of motion and is proportional to the rate of change of momentum of the upper body part.

Conversely we lean forward while the speed decreases and the same phenomenon happens in the opposite direction.

While changing direction in car the upper body remains in its position due to inertia but the lower body being firmly in contact with the car gets along in the direction of the car, seems that it makes the upper body lean in the opposite direction of the turn.

On abrupt change in the state of motion the force experienced is also intense in accordance with the Newton's second law of motion.

7 0

3 years ago

How are frequency and wave period related?

dedylja [7]

-- The unit of frequency is "per second" (Hz), which is [reciprocal time].

-- The unit of period is "second", which is [time].

Do you see where this is going ?

'Frequency' and 'period' are reciprocals of each other.

For any wave ...

Period = (1) / (frequency) .

Frequency = (1) / (period) .

6 0

3 years ago

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

$x=ut$

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

$t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s$

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

$y=\frac{1}{2} gt^2$

Substitute 9.8 m/s² for g and 0.461 s for t.

$y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m$

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by, $(2.0 m)-(1.04 m)=0.96 m$

The catcher needs to hold his glove at a height of 0,96 m from the ground.

8 0

3 years ago

The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. F

photoshop1234 [79]

The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

T - 25 N = (8 kg) a

where T is the tension in the rope and a is the acceleration.

The hanging mass has a net force of

(6 kg) g - T = (6 kg) a

where g = 9.8 m/s².

Adding these equations together eliminates T, and we can solve for a :

(T - 25 N) + ((6 kg) g - T ) = (14 kg) a

33.8 N = (14 kg) a

a = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

T - 25 N = (8 kg) (2.4 m/s²)

T ≈ 25 N + 19.31 N ≈ 44 N

5 0

3 years ago

A child and sled with a combined mass of 50.0 kg slide down a frictionless slope. if the sled starts from rest and has a speed o

Furkat [3]

Using conservation of energy

Potential Energy (Before) = Kinetic Energy (After)

mgh = 0.5mv^2

divide both sides by m

gh = 0.5v^2

h = (0.5V^2)/g

h = (0.5*2.2^2)/9.81

h = 0.25m

4 0

3 years ago

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