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VMariaS [17]
3 years ago
9

A car accelerates from rest at -3.00m/s^2. What is the velocity at the end of 5.0s? What is the displacement after5.0s?

Physics
1 answer:
Andrew [12]3 years ago
5 0

Speed = (acceleration) x (time)
Velocity = (speed) in (direction of the speed)

Speed = (-3 m/s²) x (5 s) = 15 m/s
Velocity =
             (15 m/s) in the direction opposite to the direction you call positive
.

Displacement = (distance between start-point and end-point)
                           in the direction from start-point to end-point.

Distance = (1/2) (acceleration) (time)²
Distance = (1/2) (3 m/s²) (5 s)²
                 = (1/2) (3 m/s²) (25 s²)  =  37.5 meters

Displacement =
                     37.5 meters in the direction opposite to the direction you call positive.

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A truck with 0.410 m radius tires travels at 25.0 m/s. What is the angular velocity of the rotating tires in radians per second?
Eduardwww [97]

relation between linear velocity and angular velocity is given as

v = R\omega

here

v = linear speed

R = radius

\omega = angular speed

now plug in all data in the equation

25.0 = 0.410 \omega

\omega  = \frac{25}{0.410}

\omega = 60.9 rad/s

so rotating speed is 60.9 rad/s

6 0
2 years ago
Determine the image distance and image height for a 5.00 cm tall object placed 30.0 cm from a double convex lens with a focal le
vlabodo [156]

Answer:

The image distance is 30 cm

image height = - 5 cm

Explanation:

The formula for calculating the image distance is expressed as

1/f = 1/u + 1/v

where

f is the focal length

u is the object distance

v is the image distance

From the information given,

u = 30

f = 15

By substituting these values into the formula,

1/15 = 1/30 + 1/v

1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30

Taking the reciprocal of both sides,

v = 30

The image distance is 30 cm

magnification = image height/object height = - v/u

Given that object height = 5 cm, then

image height/5 = - 30/30 = - 1

image height = - 5 * 1

image height = - 5 cm

8 0
1 year ago
The specification limits for a product are 8 cm and 10 cm. A process that produces the product has a mean of 9.5 cm and a standa
My name is Ann [436]

Answer:

The value of Cpk is 0.83.

Explanation:

Given that,

Upper specification limits = 10 cm

lower specification limits = 8 cm

Mean = 9.5

Standard deviation = 0.2 cm

We need to calculate the process capability

Using formula of Cpk

Cpk=min(\dfrac{USL-mean}{3\times SD}, \dfrac{mean-LSL}{3\times SD})

Put the value into the formula

Cpk=min(\dfrac{10-9.5}{3\times0.2}, \dfrac{9.5-8}{3\times0.2})

Cpk=min(0.83,2.5)

Cpk=0.83

Hence, The value of Cpk is 0.83.

4 0
2 years ago
According to Newton’s second law of motion what is force equals to
Maksim231197 [3]

Answer:force equals to rate of change of momentum

Explanation:

F=force

t=time

m=mass

v=final velocity

u=initial velocity

(mv-mu)/t=rate of change of momentum

Force=rate of change of momentum

F=(mv-mu)/t

8 0
3 years ago
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Question 12 (1 point) Question 12 Unsaved
Bezzdna [24]

Slope is your answer

7 0
3 years ago
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