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Lemur [1.5K]
1 year ago
7

A car with a mass of 1000 kg accelerates from 0 to 90 km/h in 10 seconds. a) What is its acceleration? b) What is the net force

on the car?
Physics
1 answer:
kompoz [17]1 year ago
3 0

Answer:

v

f=90km/h=25m/s

v_{_{i}}=0v

i=0

t=10\;\mathrm{s}t=10s

m=1000\;\mathrm{kg}m=1000kg

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A soccer player runs with a speed of 4.6m/s how long does it take him to run 60m
Misha Larkins [42]

To solve this problem divide 60 by 4.6

The answer to this problem is 13 seconds.

3 0
3 years ago
Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el
lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}

The electrostatic force between them is

F_{e}=\frac{Kq^{2}}{d^{2}}

According to the question

1 % of Fg = Fe

0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}

2.927 \times 10^{35}=9 \times10^{9}q^{2}

3.25 \times 10^{25}=q^{2}

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0
3 years ago
how to A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5
Alexandra [31]

Answer: a) 12857.1 m/s/s b) 578.6 N

Explanation:

Impulse = change in momentum

Ft = mV2 - mV1

V = AT, 45 / .0035 = 12857.1 m/s/s

(b) .045 x 12857.1 = 578.6 N

4 0
3 years ago
Read 2 more answers
A neutron consists of one "up" quark of charge +2e/3 and two "down" quarks each having charge -e/3. If we assume that the down q
Anon25 [30]

Answer:

The magnitude of the electrostatic force is 120.85 N

Explanation:

We can use Coulomb's law to find the electrostatic force between the down quarks.

In scalar form, Coulomb's law states that for charges q_1 and q_2 separated by a distance d, the magnitude of the electrostatic force F between them is:

F = k \frac{|q_1q_2|}{d^2}

where k is Coulomb's constant.

Taking the values:

d = 4.6 \ 10^{-15} m

q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}

and knowing the value of the Coulomb's constant:

k = 8.99 \ 10 ^{9} \frac{N m^2}{C^2}

Taking all this in consideration:

F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}

F = 120.85  \ N

8 0
3 years ago
A cart with a mass of 0.5 kg is at the top of the ramp. The height is 0.40m .
Tju [1.3M]

A=0.05.0M.

B=68.9244GPE.34

C=0

D it would be 79%HIGHER

3 0
3 years ago
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