Hi Pupil Here is your answer ::
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1 The shape of the Body
Example : The shape of the ball lying on a floor can be changed by pressing it.
2 Direction of the Body
Example : The direction of motion of moving ball can be changed by hitting it with a bat.
3 The speed of the Body
Example : A ball at rest can be set in motion if force is applied only
4. Size of the Body
Example : The length of a spring tied and on one end can be increased by pulling it.
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Hope this helps .......
Answer: C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.
Explanation:
The speed of each car is defined as:
where d is the distance traveled by the car and t is the time taken.
For the yellow car, d=400 mi and t=8 h, so its speed is
For the green car, d=400 mi and t=10 h, so its speed is
So, the correct choice is
C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m
Answer:
the boat would be deeped by 3200 m
Explanation:
Given that
The boat arrives back after 4 seconds
And, the speed of the sound in water is 1,600 m/s
We need to find out how much deep is the water
So,
As we know that
Distance = ( speed × time) ÷ 2
Here we divided by 2 because the boat arrives back
= (1600 × 4) ÷ 2
= 3200 m
Therefore the boat would be deeped by 3200 m
