1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lozanna [386]
2 years ago
6

A ball is thrown upward with an initial velocity of 12.1 m/s. How long will it take to reach a velocity of -24.5 m/s?

Physics
1 answer:
Rainbow [258]2 years ago
5 0
<h3><u>Answer;</u></h3>

 = 3.73 seconds

<h3><u>Explanation;</u></h3>

Using the equation;

v = u + at

where v is the final velocity, u is the initial velocity and a is the acceleration due to gravity (-g) and t is the time taken.

Therefore;

v = u - gt

Thus;

v = 24.5 m/s, u = 12.1 m/s, a = -g = 9.8 m/s²

-24.5 m/s = 12.1 - 9.8 × t

36.6 = 9.8 t

t = 36.6/9.8

 = 3.73 seconds

You might be interested in
A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back
castortr0y [4]

Answer:

a) 5 m/s downwards

b) 17.86 m/s

c) 24.82 m/s

d) 0.228

Explanation:

We can set the frame of reference with the origin on the top of the building and the X axis pointing down.

The rock will be subject to the acceleration of gravity. We can use the equation for position under constant acceleration and speed under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

V(t) = V0 + a * t

In this case

X0 = 0

V0 = -5 m/s

a = 9.81 m/s^2

To know the speed it will have when it falls back past the original point we need to know when it will do it. When it does X will be 0.

0 = -5 * t + 1/2 * 9.81 * t^2

0 = t * (-5 + 4.9 * t)

One of the solutions is t = 0, but this is when the rock was thrown.

0 = -5 + 4.69 * t

4.9 * t = 5

t = 5 / 4.9

t = 1.02 s

Replacing this in the speed equation:

V(1.02) = -5 + 9.81 * 1.02 = 5 m/s (this is speed downwards because the X axis points down)

When the rock is at 15 m above the street it is 15 m under the top of the building.

15 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 15 = 0

Solving electronically:

t = 2.33 s

At that time the speed will be:

V(2.33) = -5 + 9.81 * 2.33 = 17.86 m/s

When the rock is about to reach the ground it is at 30 m under the top of the building:

30 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 30 = 0

Solving electronically:

t = 3.04 s

At this time it has a speed of:

V(3.04) = -5 + 9.81 * 3.04 = 24.82 m/s

---------------------

Power is work done per unit of time.

The work in this case is:

L = Ff * d

With Ff being the friction force, this is related to weight

Ff = μ * m * g

μ: is the coefficient of friction

L = μ * m * g * d

P = L/Δt

P = (μ * m * g * d)/Δt

Rearranging:

μ = (P * Δt) / (m * g * d)

1 horsepower is 746 W

20 minutes is 1200 s

μ = (746 * 1200) / (100 * 9.81 * 4000) = 0.228

8 0
2 years ago
on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker
Aleksandr-060686 [28]
B4 the tackle: 

<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>

<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>


<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>

<span>The vector triangle is right angled: </span>

<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>

<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>

<span>v(f) = 5.6 m/s (to 2 sig figs) </span>


<span>direction of v(f) is the same as the direction of the final momentum </span>

<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>


<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>




<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
4 0
3 years ago
Which planetery body has the greast gravitational pull?
Nataly [62]
Answer should be the earth
7 0
2 years ago
Answers are - <br><br>A 235 N<br><br>B 376 N<br><br>C 271 N<br><br>D 188 N<br><br>E 470 N
amm1812

Answer:

C

Explanation:

8 0
3 years ago
Why are such scientific advances still valuable?
Sergeu [11.5K]
Because the more advances made in the world means the more we can learn on how things work and how we can better the lives of humans and other species. If we didn't have scientific advancements we wouldn't have cell phones, electric, tv, car, computers, ect. We would still be living in Cave man era with clubs and horrible language skills.
4 0
3 years ago
Other questions:
  • Two objects that may be considered point masses are initially separated by a distance d. The separation distance is then decreas
    9·1 answer
  • 4 meters and the frequency is 3 hz what’s the wave speed
    5·2 answers
  • A wheel is rotating at 30.0 rpm. The wheel then accelerates uniformly to 50.0 rpm in 10.0 seconds. Determine the – a) rate of an
    15·2 answers
  • The convergence of two continental plates would produce
    11·1 answer
  • The sun is 60° above the horizon. Rays from the sun strike the still surface of a pond and cast a shadow of a stick that is stuc
    15·1 answer
  • If a net force of 25 N is exerted over a distance of 4 m to the right on a 2 kg mass initially at rest and moves it, what is the
    11·1 answer
  • 1) When making a digital animation of a person running on a sidewalk in a scene, which parameter would be an initial condition?
    10·1 answer
  • An eagle is flying horizontally at a speed of 2.60 m/s when the fish in her talons wiggles loose and falls into the lake 4.70 m
    15·2 answers
  • A candy-filled piñata is hung from a tree for Matthew's birthday. During an unsuccessful attempt to break the 4.4-kg piñata, Hay
    13·1 answer
  • A car starting from rest accelerates in a straight line at a constant rate of 5.5m/s for 6s.If the car after this acceleration s
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!