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dybincka [34]
4 years ago
15

35. A daredevil is launched out of a

Physics
1 answer:
Nezavi [6.7K]4 years ago
6 0

Answer:

845 m

Explanation:

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 13 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (13 s) + ½ (10 m/s²) (13 s)²

Δy = 845 m

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the force that is exerted when a shopping cart is pushed. the forces that causes a metal ball to move toward a magnet
alexandr402 [8]

Answer:

The force that is exerted when a shopping cart is pushed is a type of push force, supplied by the muscles of the cart pusher's body.

The forces that causes a metal ball to move toward a magnet is a type of pull force that is as a result of the magnetic field forces.

Explanation:

Forces are divided into push forces that tends to accelerate a body away from the source of the force, and pull forces that accelerates the body towards the force source.

Examples of push forces includes pushing a cart, pushing a table, repulsion of two similar poles of a magnet etc. Examples of pull forces includes a attractive force between two dissimilar poles of a magnet, pulling a load by a rope, a dog pulling on a leash etc.

8 0
3 years ago
Read 2 more answers
In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp
MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

V_{Ax}=(21m/s)cos(-14^{o})

V_{Ax}=20.38 m/s

V_{Ay}=V_{A}sin\theta

V_{Ay}=(21m/s)sin(-14^{o})

V_{Ay}=-5.08 m/s

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0
3 years ago
Applications of pressure
Sunny_sXe [5.5K]
  • hydraulic press
  • hydraulic lift
  • hydraulic jack
  • hydraulic brake
3 0
2 years ago
A billiard ball (ball #1) moving at 5.00 m/s strikes a stationary ball (ball #2) of the same mass. after the collision, ball #1
Rashid [163]
If it is completely elastic, you can calculate the velocity of the second ball from the kinetic energy 
<span>v1 = velocity of #1 </span>
<span>v1' = velocity of #1 after collision </span>
<span>v2' = velocity of #2 after collision. </span>

<span>kinetic energy: v1^2 = v1' ^2 + v2' ^2 (1/2 and m cancel out) </span>
<span>5^2 = 4.35^2 + v2' ^2 </span>
<span>v2 = 2.46 m/s <--- ANSWER</span>
4 0
3 years ago
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Balances are used to measure what a. Mass of an object b. Volume c. Force acting upon it d. Level surfaces on the table
suter [353]
<span>a. Mass of an object </span>
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3 years ago
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