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Zigmanuir [339]
3 years ago
7

A car accelerates forward at 3.5 m/s2. If the car has a mass of 873 kg, what is the force applied to the car?

Physics
2 answers:
just olya [345]3 years ago
4 0
3055. 5 kg*m/s2 you do (873kg)(3. 5m/s2)
Ierofanga [76]3 years ago
3 0

Answer:

3055.5 N

Explanation:

a = 3.5 m/s^2, m = 873 kg

According to the Newton's second law of motion, force acting on a body is the product of mass of body and the acceleration of body.

F = m x a = 873 x 3.5 = 3055.5 N

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What is the name of the british scientist credited with discovering cathode rays?
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3 years ago
Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the
Doss [256]

Answer:

v_{su} = 19.44 m/s

Explanation:

m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg

T=9.29x10^8\\r_{o}=1.43x10^{12}

If the sun considered as x=0 on the axis to put the center of the mass as a:

m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}

solve to r1

r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}

r_1=1.428x10^9m

Now convert to coordinates centered on the center of mass.  call the new coordinates x' and y' (we won't need y').  Now since in the sun centered coordinates the angular momentum was  

L = \frac{m_{sa}*2*pi*r_1^2}{T}

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum.  So that means

L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}

Since

L_{su}= m_{su}*v_{su}*r_1

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v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}

v_{su} = 19.44 m/s

7 0
3 years ago
A satellite orbits the earth at a speed of 8.0x 10^3m/s . Calculate the period of orbit of the satellite if the distance between
gladu [14]

Hi there!

The period of an orbit can be found by:

T = \frac{2\pi r}{v}

T = Period (? s)
r = radius of orbit (6400000 m)

v = speed of the satellite (8000 m/s)

This is the same as the distance = vt equation. The total distance traveled by the satellite is the circumference of its circular orbit.

Let's plug in what we know and solve.

T = \frac{2\pi (6400000)}{8000} = \boxed{5026.55 s}

8 0
1 year ago
A vacuum tube can be used to__. A. change alternating current into direct current B. increase the strength of a signal.. C. turn
vesna_86 [32]

The correct answer of this question is :  A) Change alternating current into direct current.

EXPLANATION :

As per the question, we are given vacuum tube. Vacuum tube can be of various types. Normally it contains two electrodes called cathode and anode which are enclosed in an evacuated glass chamber . There are also other types of vacuum tubes which contain extra electrodes like control grid .

The vacuum tube can be used as a rectifier. It means that it can be used as an electronic device which will convert alternating current into direct current. It may be a half wave rectifier or a full wave rectifier. Actually the direct current obtained during the rectification of alternating current is pulsating in nature.

Hence, the correct answer is that a vacuum tube can be used to change alternating current into direct current.


4 0
3 years ago
Read 2 more answers
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