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Sunny_sXe [5.5K]
1 year ago
5

How can you increase the force required to move an object without changing the mass of the object?

Physics
1 answer:
Effectus [21]1 year ago
8 0

To increase the force required to move an object without change mass is : To increase the acceleration of the object

<h3>Force acting on an object </h3>

Given that Force = Mass * acceleration

An increase in the acceleration of an object in motion will result in a proportional increase in the Force required to move the object becasue force of an object is directly proportional to the acceleration of an object.

Hence we can conclude that To increase the force required to move an object without change mass is To increase the acceleration of the object.

Learn more about Force : brainly.com/question/25239010

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The separation in time between the arrival of primary and secondary waves is
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The separation in time between the arrival of primary and secondary wave is called LAG TIME.
The time difference between the arrival of primary wave and secondary wave in a seismogram is called lag time. The primary wave always travels faster than the secondary wave, thus the difference between the two can be obtained by estimating the difference between the arrival time of the two waves/.
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3 years ago
Which statement is supported by this scenario?
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Answer:

For Yanni, the speed of the ball is 15 m/s, and for the quarterback, the speed of the ball is 8 m/s.

Explanation:

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2 years ago
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A force vector F1 points due east and has a magnitude of 200N. A second force F2 is added to F1. The resultant of the two vector
PilotLPTM [1.2K]

Answer:

The second vector \vec{F_2} points due West with a magnitude of 600N

Explanation:

The original vector \vec{F_1} points with a magnitude of 200N due east, the Resultant vector \vec{R} points due west (that's how east/west direction can be interpreted, from east to west) with a magnitude of  400N. If we choose East as the positive direction and West as the negative one, we can write the following vectorial equation:

\vec{F_1}+\vec{F_2}=\vec{R}\implies\vec{F_2}=\vec{R}-\vec{F_1}=-400N-200N=-600N

With the negative sign signifying that the vector points west.

3 0
3 years ago
A skydiver has jumped out of a plane and is falling faster and faster. what forces are present in this situation
kompoz [17]
Gravity and air resistance 

i took the test and got 100%
5 0
3 years ago
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
3 years ago
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