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Delicious77 [7]
3 years ago
14

A car left point a at 7:30 am and arrived at point b, 162 miles away at 10:30 am. what was its average speed in miles per hour?5

3545556skip
Physics
1 answer:
Reika [66]3 years ago
3 0
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A 20-N force acts on an object with a mass of 2.0kg. what is the objects acceleration?
ale4655 [162]

Answer:

10 m/s^2

Explanation:

Equation: F = ma.

a = acceleration

m = mass

F = force

Because we are trying to find acceleration instead of force we want to rearrange the equation to solve for a which is F/m = a.

F = 20

m = 2

a = ?

a = F/m

a = 20/2

a = 10 m/s^2

7 0
3 years ago
Read 2 more answers
On a nice summer day,Kim takes her niece Madison for a walk in her stroller.If they start from rest and accelerate at a rate of
11111nata11111 [884]

2.5m/s

Explanation:

Given parameters:

Initial velocity = 0m/s

Acceleration = 0.5m/s²

time of travel = 5s

Solution:

Final velocity = ?

Solution:

Acceleration can be defined as the change in velocity with time:

          Acceleration = \frac{Final velocity - Initial velocity}{time}

  From the equation above, the unknown is final velocity:

Final velocity - initial velocity = Acceleration x time

 since initial velocity = 0

   Final velocity = 0.5 x 5 = 2.5m/s

Learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

6 0
3 years ago
A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0
2 years ago
2. Is this chemical equation balanced?<br> 2C4H10 (g) + 13O2 (g) = 8CO2 (g) → 10H2O (l)
ehidna [41]
Yes that is a balaned equation
3 0
3 years ago
If you're studying the science of sound, you're studying____?
timurjin [86]
The study of sound is called sonics and the study of sound waves are acoustics
3 0
3 years ago
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