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3 years ago

What must happen to an object in order to accelerate it?

Physics

1 answer:

White raven [17]3 years ago

6 0

Answer:

For an object to accelerate, appropriate force must be applied to the object to cause it to change it’s velocity if it’s already in motion. However, to cause it to overcome static friction if it is at rest, and cause it to change it’s velocity

Explanation:

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If the ice cube absorbs heat, then, heat

Rashid [163]

Answer:

<h2><em>I have no idea what any of that means </em>:'(</h2>

Explanation:

<h2><em>So Sorry. ΓΟΛ3</em></h2>

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3 years ago

HURRY! ILL MARK YOU AS BRAINLIEST, RATE YOU A 5 AND THANK YOU! 50 POINTS!

nignag [31]

Answer:

The correct answer would be answer A

Explanation:

It has to be A because if the seeds are hetrozygous there is a chance that if u make a punnet square that u can get a smooth ressesive seed

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4 years ago

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A large ball of Play-Doh with a mass of 0.04kg is launched from a catapult with an initial

ryzh [129]

Answer:

≈ 6.68 m/s

Explanation:

A suitable formula is ...

vf^2 -vi^2 = 2ad

where vi and vf are the initial and final velocities, a is the acceleration, and d is the distance covered.

We note that if the initial launch direction is upward, the velocity of the ball when it comes back to its initial position is the same speed, but in the downward direction. Hence the problem is no different than if the ball were initially launched downward.

Then ...

vf = √(2ad +vi^2) = √(2·9.8 m/s^2·1.0 m+(5 m/s)^2) = √44.6 m/s

vf ≈ 6.68 m/s

The ball hits the ground with a speed of about 6.68 meters per second.

We assume the launch direction is either up or down.

3 0

3 years ago

assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves th

finlep [7]

Answer:

A. xmax = 131.49 m

B. t = 8.74 s

C. ymax = 220.33 m

Explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

$y=y_o+v_osin\theta-\frac{1}{2}gt^2$ (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

$0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2$ (2)

You use the quadratic formula to obtain the value of t:

$t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s$

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

$x_{max}=v_ocos\theta t$

$x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m$

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

$y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}$ (3)

By replacing in the equation (3) the values of all parameters you obtain:

$y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m$

The maximum height reached by the cannon ball is 220.33m

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3 years ago

Which of the following are mechanical

Natasha2012 [34]

Answer:

Hey how are you today

Your answer would be B sound and ocean waves.

The reason why of course is because mechanical waves are waves that travel through matter and not empty space (as for light travels through empty space and is therefore not a machanical wave). and x-rays are a spectrum of light so they aren't mechanical waves eather.

Hope it helped

6 0

3 years ago

Read 2 more answers