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ZanzabumX [31]
3 years ago
15

If you swing an object on the end of a string around a circle, the string pulls on the object to keep it moving in a circle. Wha

t is the name of this force?
A. inertial
B. centripetal
C. resistance
D. gravitational
Physics
1 answer:
emmainna [20.7K]3 years ago
4 0

Answer:

B

Explanation:

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The 1.5kg ball is launched straight upward with an initial velocity of 7m/s. What is the maximum height it will reach?​
Temka [501]

Answer:

h = 2.5 m

Explanation:

Given that,

Mass of a ball, m = 1.5 kg

Initial velocity of the ball, u = 7 m/s

We need to find the maximum height reached by the ball. Let it is be h. Using the conservation of energy to find it such that,

mgh=\dfrac{1}{2}mu^2\\\\h=\dfrac{u^2}{2g}

Put all the values,

h=\dfrac{7^2}{2\times 9.8}\\\\=2.5\ m

So, it will reach to a height of 2.5 m.

8 0
3 years ago
From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the
Ivahew [28]

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

6 0
3 years ago
Please Help! Im dumb.
Rus_ich [418]
The answers false I believe
8 0
3 years ago
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Why is it wrong to leave our light on​
Dennis_Churaev [7]

Answer:

you will get huge electricity bills ............

8 0
3 years ago
People with good hearing can perceive sounds as low in level as −7.53 dB at a frequency of 3000 Hz. What is the intensity of thi
lora16 [44]

Answer:

attached below

Explanation:

3 0
3 years ago
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