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Anestetic [448]
3 years ago
15

HELP PLEASE! write the equation for calculating relative velocity in one dimensional.

Physics
1 answer:
RoseWind [281]3 years ago
6 0

Anwer:

Key Equations

Position Vector: `n r (t) = (t)^j+y

Total acceleration: `n a = `n aC

+ `n aT

Position Vector in frame S is the position Vector in frame S` plus the vector from the origin of S to the origin of S`: `n rPS = `n rPS`

+ `n rS`S

Relative velocity equation connecting two references frame: `n vPS = `n vPS`

+ `n vS`S

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A man rides up in an elevator at 12 m. He gains 6500 J of gravitational potential energy. what is the man's mass?
iren2701 [21]

Answer:

We know that potential energy of a body;

= mass(m)× gravitational acceleration(g) × height(h)

Lets find out the mass of the body

P.E. = mgh

=> 6500J = mass × 9.8m/s^2 × 12m

=>6500J = mass × ( 9.8 × 12 ) × ( m/s^2 × m)

=> 6500 Nm = m × 117.6 × m^2 / s^2

=> 6500/117.6 Ns^2/m = mass [°.° Ns^2/m = kg]

=> 55.272 Kg = mass

Therefore the mass of the body = 55.272 kg ~ <em>6</em><em>0</em><em> </em><em>k</em><em>g</em><em> </em>(Ans)

Hope it helps you

6 0
3 years ago
Which type of training will improve muscular strength?
mojhsa [17]

Answer:

C

Explanation:

Lifting heavy weights every day

5 0
3 years ago
The fulcrum of a uniform 20-kg seesaw that is 4.0 m long is located 2.5 m from one end. A 30-kg child sits on the long end. Part
uranmaximum [27]

Answer:

57 kg

Explanation:

Mass of seesaw = 20 kg

Length of seesaw = 4 m

Mass of child on the longer end = 30 kg

The weight of the seesaw acts at the center i.e. 2m

The algebraic sum of moments of all forces about any point is zero, hence, using the fulcrum as the reference point:

[x * 9.8* 1.5] - [20 * 9.8* (2.5 - 2)] - [30 * 9.8 * 2.5] = 0

=> 14.7x = (20*9.8*0.5) + 735

14.7x = 98 + 735

14.7x = 833

=> x = 833/14.7

x = 57 kg

7 0
3 years ago
An object has a momentum of 55kg*m/s and hits a stationary object making the second object starts to move. If the first object e
Afina-wow [57]

Let the moving object be object 1 and the stationary object be object 2.

Momentum of object 1 before collision = 55 kg*m/s

Momentum of object 2 before collision = 0 kg*m/s

Momentum of object 1 after collision = 13 kg*m/s

According to the law of conservation of momentum, the sum of the momenta of 2 objects remains the same even after collision. So,

Momentum of object 1 before collision + momentum of object 2 before collision = Momentum of object 1 after collision + momentum of object 2 after collision

55 + 0 = 13 + momentum of object 2 after collision

Momentum of object 2 after collision = 55 - 13 = 42 kg*m/s

Hope I helped!

4 0
3 years ago
If I have an average speed of 50 km/hr and I travel 150 km , how long did the trip take me ?
Ludmilka [50]
150 km / 50 km/hr = 3 hr
4 0
3 years ago
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