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dangina [55]
3 years ago
5

A ______ is a push or pull that one object exerts on another.

Physics
2 answers:
fiasKO [112]3 years ago
8 0
The answer is A. force. 
lilavasa [31]3 years ago
6 0
A. Force is  push or pull.
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How often does a 16-year-old need a medical evaluation from a doctor?
a_sh-v [17]

Answer:

Explanation: A

5 0
4 years ago
Read 2 more answers
If Scobie could drive a Jetson's flying car at a constant speed of 160.0 km/hr across oceans and space, approximately how long w
DochEvi [55]

Scobie will take 10 days to drive around Earth's equator.

To calculate the time that takes Scobie to drive around Earth's equator we need to find the distance, which is given by the equation of a circumference:

d = 2\pi r

<em>Where:</em>

r: is the Earth's radius = 6371 km

Then, the distance is:

d = 2\pi r = 2\pi*6371 km = 40030.2 km

Now, if we divide the above distance by the speed of the car we can find the time:

t = \frac{d}{v} = \frac{40030.2 km}{160.0 km/h} = 250.2 h*\frac{1 d}{24 h} = 10 d

Therefore, Scobie will take 10 days to drive around Earth's equator.

     

To learn more about distance and time here: brainly.com/question/14236800?referrer=searchResults

I hope it helps you!

6 0
3 years ago
A person pulls a bucket of water up from a well with a rope. Assume the initial and final speeds of the bucket are zero (Vi-Vf-0
n200080 [17]

Answer:

a

This a closed system because the mass of the system is conserved

The energy system that undergoes change is the Potential energy system

The energy system diagram is shown on the first uploaded image

b

Work done = Change in gravitational potential energy

So solving algebraically for work done would be

    Work done   = m*g*h

where m is mass

          g is acceleration due to gravity

          and h is the height

c

Work done in terms of force and distance is = mg

where  m is mass of bucket and

            g is acceleration due to gravity  

Explanation:

a) At the start, potential and kinetic energy were zero. so, energy is zero.

As the person pulls the bucket up, the potential energy becomes mgh.

so,final energy will be consisting of only potential energy.

B) Here work done is equal to change in gravitational potential energy.

W = \Delta P.E

W = m*g*h

where g = 9.9 m/s^2

C) Work = force * distance

mgh = force * h

force = mg

force = weight of bucket

6 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
3 years ago
A 5 newton force and a 7 newton force act concurrently on a point. As the angle between the forces is increased from 0 to 180 th
Reika [66]

Answer:

The magnitude of the resultant decreases from A+B to A-B

Explanation:

The magnitude of the resultant of two vectors is given by

R=\sqrt{A^2 +B^2 +2AB cos \theta}

where

A is the magnitude of the first vector

B is the magnitude of the second vector

\theta is the angle between the directions of the two vectors

In the formula, A and B are constant, so the behaviour depends only on the function cos \theta. The value of cos \theta are:

- 1 (maximum) when the angle is 0, so the magnitude of the resultant in this case is

R=\sqrt{A^2 +B^2+2AB}=\sqrt{(A+B)^2}=A+B

- then it decreases, until it becomes 0 when the angle is 90 degrees, where the magnitude of the resultant is

R=\sqrt{A^2 +B^2+0}=\sqrt{A^2+B^2}

- then it becomes negative, and continues to decrease, until it reaches a value of -1 when the angle is 180 degrees, and the magnitude of the resultant is

R=\sqrt{A^2 +B^2-2AB}=\sqrt{(A-B)^2}=A-B


4 0
3 years ago
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