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Paraphin [41]
3 years ago
8

I need help with this too. (im not good at science or math)

Physics
2 answers:
dedylja [7]3 years ago
7 0
I believe 1 m/s.....
mr_godi [17]3 years ago
6 0

Answer:

i belive 1 m/s

Explanation:

dividing displacement from time it should be 1 cuz 5/5 is 1

please tell me if right!

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7. What is the velocity of an object with a distance of 90m south and a time of<br> 5s?
IrinaK [193]

Answer:

Explanation:

v= s/t

V =90m/5s

V = 8m/s

4 0
2 years ago
what is the mass of a cannon ball if a force 2500 N gives the cannon ball an acceleration of 200 m/s squared?
nignag [31]

Using Newton's second law of motion:

F=ma ;  [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)


Given:                      Find:                   Formula:                  Solve for m:

F: 2500N                 mass:?                F=ma Eq.1              m=F/a  Eq. 2

a= 200m/s^2  


Solution:

Using Eq.2

m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg

8 0
3 years ago
a block with length 1.5m width 1m height 0.5m and mass 300kg lays on the table.what is the pressure at the bottom surface of the
Nesterboy [21]

Answer:

your answer will be 320kg that would be the pressure at the bottom surface of the block

6 0
2 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
A colloidal liquid in a has is called
S_A_V [24]
Aerosol is the answer
8 0
3 years ago
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