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Luba_88 [7]
1 year ago
14

I am giving a reward of 17 points do you want more rewards?

Physics
1 answer:
stiks02 [169]1 year ago
4 0

Answer: I do not mind if I have more rewards or not.

Explanation: I think helping others is the best reward

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What is the magnitude of velocity for a 3,100 kg car possessing 4,100 kg•m/s of momentum?​
azamat

Answer:

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Explanation:

3 0
3 years ago
Read 2 more answers
How much force is needed to accelerate a 1,800kg car at rate of 1.5 m/s2?
topjm [15]

<u>Given data:</u>

acceleration (a) = 1.5 m/s² ,

           mass (m) = 1800 Kg ,

        Determine F = ?

         <em>From Newtons II law</em>

                          F = m.a   N

                              = 1800× 1.5

                             = 2700 N

<em>2700 N force needed to accelerate the car</em>

7 0
3 years ago
Two positive charges of 6.0 x 10^-6 C are separated by 0.50 m. Calculate the force of electrical attraction
SpyIntel [72]

Answer:

Force between two charges is given as 1.296 N

Explanation:

As we know that the two charges will attract or repel each other when they are placed near to each other

Here we know that electrostatic force is given as

F = \frac{kq_1 q_2}{r^2}

here we have

q_1 = q_2 = 6 \times 10^{-6} C

distance between two charges is given as

r = 0.50 m

so we have

F = \frac{(9\times 10^9)(6 \times 10^{-6})^2}{0.50^2}

F = 1.296 N

3 0
3 years ago
What units are used to measure force
Nastasia [14]
N, or Neutons are a common way to measure force
5 0
4 years ago
Read 2 more answers
A 65-cm segment of conducting wire carries a current of 0.35
algol13
The intensity of the magnetic force exerted on the wire due to the presence of the magnetic field is given by
F=ILB \sin \theta
where
I is the current in the wire
L is the length of the wire
B is the magnetic field intensity
\theta is the angle between the direction of the wire and the magnetic field

In our problem, L=65 cm=0.65 m, I=0.35 A and B=1.24 T. The force on the wire is F=0.26 N, therefore we can rearrange the equation to find the sine of the angle:
\sin \theta= \frac{F}{ILB}= \frac{0.26 N}{(0.35 A)(0.65 m)(1.24 T)}=0.922

and so, the angle is
\theta=\arcsin(0.921)=67.1^{\circ}
6 0
3 years ago
Read 2 more answers
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