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zheka24 [161]
3 years ago
7

A box of mass 5.0 kg is initially at rest. A 500 N force causes it to move horizontally through a

Physics
2 answers:
ANEK [815]3 years ago
7 0

Explanation:

Work done= 500 × 12= 6000J=6KJ

HOPE it helps

me too bored

melamori03 [73]3 years ago
4 0

can u give me brainliest and I'll talk with u

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Why do objects repel and attract?
Anarel [89]
Objects repel and attract because of a thing called electrostatic attraction. When objects have the same charge (positive or negative), then they will repel, and if they have opposite charges then they will attract
5 0
3 years ago
A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught
meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2} (Eq. 1)

Where:

y_{o} - Initial height of the softball, measured in meters.

y - Final height of the softball, measured in meters.

v_{o} - Initial velocity of the softball, measured in meters per second.

t - Time, measured in seconds.

g - Gravitational acceleration, measured in meters per square second.

If we know that y = y_{o}, t = 3.56\,s and g = -9.807\,\frac{m}{s^{2}}, the initial velocity of the softball is:

v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0

3\cdot v_{o} -44.132\,m= 0

v_{o} = 14.711\,\frac{m}{s}

The initial velocity of the softball is 14.711 meters per second.

8 0
3 years ago
At what angle are the electronic and the magnetic wave related in an electromagnetic signal?
VikaD [51]

Answer:

90degrees I'm pretty sure

7 0
3 years ago
A curve in a stretch of highway has radius R. The road is not banked in any way. The coefficient of static friction between the
adelina 88 [10]

Answer:

maximum possible velocity = \sqrt{ugR}

Explanation:

centripetal acceleration when the  car is going in the circle must be less than the maximum friction for the car to not slip.

centripetal acceleration \frac{mv^{2}}{r}

where v is the velocity of car and r is the radius of circle

maximum friction = umg

where u is the coefficient of static friction.

thereforeumg\geq \frac{mv^{2}}{R}

therefore maximum possible velocity = \sqrt{ugR}

6 0
3 years ago
Identify the three types of isometric transformations. A. reflection, rotation, translation B. reflection, rotation, dilation C.
inessss [21]
<span>reflection, rotation, translation</span>
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3 years ago
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