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s2008m [1.1K]
3 years ago
7

At what angle should an object be thrown so that the maximum range is equal to three times the height reached by the projectile?

Physics
1 answer:
Delvig [45]3 years ago
5 0
This link might help you: https://www.quora.com/At-what-angle-to-the-horizontal-should-a-body-be-projected-so-that-maximum-height-reached-is-equal-its-horizontal-range
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The electrons equal or copy the number of _____
Westkost [7]
Protons copy the number equal
3 0
3 years ago
Read 2 more answers
A car goes from 5 m/s to 25 m/s in 6 s. What is the acceleration of the car?
damaskus [11]

Answer:

5m/s

Explanation:

3 0
3 years ago
Describe a ball's motion as it rolls up a slanted
emmasim [6.3K]

The ball will decelerate as it moves upwards.

The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

The given parameters;

  • initial velocity of the ball, u = 1.25 m/s
  • time of motion of the ball, t = 4.22 s

As the ball rolls up the inclined plane, the velocity decreases and eventually becomes zero when the ball reaches the highest point of the plane.

Thus, the ball decelerate as it moves upwards.

The acceleration of the ball is calculate as;

a = \frac{v_f -v_0}{t} \\\\

<em>at the highest point on the incline plane, the final velocity </em>v_f<em> is zero</em>

a = \frac{0-1.25}{4.22} \\\\a = -0.3 \ m/s^2

Thus, the magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

Learn more here:brainly.com/question/23860763

4 0
3 years ago
A 235 kg crate is pulled across a horizontal surface with a force of 760 N applied at an
AnnyKZ [126]

Answer:

658.16N

Explanation:

Step one:

given data

mass m= 235kg

Force F= 760N

angle= 30 degrees

Required

The horizontal component of the force

Step two:

The horizontal component of the force

Fh= 760cos∅

Fh=760cos30

Fh=760*0.8660

Fh=658.16N

3 0
3 years ago
Which statement is true for a car that first goes around a curve of radius r at a constant speed v, and then goes around the sam
sasho [114]
Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around
3 0
3 years ago
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