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4 years ago

What is the acceleration of a 25 kg mass pushed by a 10n force

Physics

2 answers:

Kamila [148]4 years ago

5 0

f divided by m = a

a = 0.4

Sidana [21]4 years ago

5 0

Answer:

The answer is acceleration is equal to 0.4m/s^2

Step by Step Explanation:

1) Acceleration is a vector derived magnitude that indicates the variation of speed per unit of time.

2) Taking into account Newton's Second Law, which determines a proportional relationship between force and variation in the amount of movement or linear momentum of a body. That is, the force is directly proportional to the mass and acceleration of a body.

3) This law is expressed by the formula F = m.a where (F) represents Force (m) represents mass and (a) represents acceleration.

4) Knowing this we can issolate (a) from the formula and we would have:

$a = \frac{F}{m}$

5) Having this we can now replaces the values given in the problem.

$a = \frac{10N}{25Kg} \\a = 0.4$

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An electron in a vacuum is first accelerated by a voltage of 81700 V and then enters a region in which there is a uniform magnet

Vilka [71]

Answer:

Magnetic force is equal to $1.37\times 10^{-11}N$

Explanation:

We have given electron is accelerated with a potential difference of 81700 volt.

Magnetic field B = 0.508 T

Angle between magnetic field and velocity $\Theta =90^{0}$

Mass of electron $m=9.11\times 10^{-31}kg$

Charge on electron $e=1.6\times 10^{-19}C$

By energy conservation.

$\frac{1}{2}mv^2=qV$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=1.6\times 10^{-19}\times 81700$

$v=169.4\times 10^6m/sec$

Magnetic force on electron

$F=qvBsin\Theta$

$F=1.6\times 10^{-19}\times 169.4\times 10^6\times 0.508\times sin90^{\circ}$

$=1.37\times 10^{-11}N$

8 0

3 years ago

Read 2 more answers

A car is initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at

Aleks04 [339]

Acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
= 37.5 m/s
velocity initial = 35 km/hr x 1hr /3600 s x 1000 m/1 km
= 9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units = (2.646m/s^2)/(9.8m/s^2)
= 0.27 units

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4 years ago

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3 years ago

The maximum tension the wire can withstand without breaking is 300 N . A 0.800 kg projectile traveling horizontally hits and emb

topjm [15]

Answer:

Let's investigate the case where the cable breaks.

Conservation of angular momentum can be used to find the speed.

$\vec{L}_1 = \vec{L}_2\\\vec{L}_1 = m\vec{v_0} \\\vec{L}_2 = I\vec{\omega}\\$

The projectile embeds itself to the ball, so they can be treated as a combined object. <u>The moment of inertia of the combined object is equal to the sum of the moment of inertia of both objects. </u>

$I = I_{projectile} + I_{ball}\\I = mr^2 + mr^2\\I = 2mr^2$

where r is the length of the cable.

<u>After the collision, the ball and the projectile makes a circular motion because of the cable.</u> So, the force (tension) in circular motion is

$F = \frac{mv^2}{r}$

The relation between linear velocity and the angular velocity is

$v = \omega r$

So,

$F = \frac{m(\omega r)^2}{r} = m\omega^2 r = 300\\mv_0r =I\omega\\\\\omega = mv_0r/I\\300 = m(\frac{mv_0r}{I})^2r = m(\frac{mv_0r}{2mr^2})^2r = m(\frac{v_0}{2r})^2r = \frac{mv_0^2r}{4r^2} = \frac{mv_0^2}{4r}\\300 = \frac{0.8v_0^2}{4r}\\1500 = v_0^2/r\\v_0 = \sqrt{1500r}$

As can be seen, the maximum velocity for the projectile without breaking the cable is $\sqrt{1500r}$ , where r is the length of the cable.

6 0

3 years ago

Speed can be calculated by dividing distance by what?

Nady [450]

S=d/t

speed can be calculated by dividing distance by time.

3 0

3 years ago

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