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Ray Of Light [21]
3 years ago
5

If an object has a mass of 38 kg, what is its approximate weight on earth?

Physics
1 answer:
klasskru [66]3 years ago
3 0
38*10=380 N
To be more exact, 38 should be multiplied by 9.8 instead of 10.
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Describe the movement of sperm in the female reproductive system. State the different structures that it must pass in order to m
saw5 [17]

Answer:

After a male ejaculates, many sperms move to the upper vagina (via contractions from the vagina) through the cervix and across the length of the uterus, reaching the fallopian tubes. Here they will meet the egg cell ready to be fertilized

5 0
2 years ago
A 1900kg car starts from rest and drives around a flat 65-m-diameter circular track. The forward force provided by the car's dri
s344n2d4d5 [400]

Answer:

The\quad magnitude\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad \quad =2.52m/{ s }^{ 2 }\\ The\quad direction\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad =15.{ 72 }^{\o}\\The\quad car\quad begins\quad to\quad slide\quad out\quad \quad of\quad the\quad circle\quad after\quad 26.09s.\quad \quad \quad \quad

Explanation:

8 0
3 years ago
A 1400 kg car traveling at 17.0 m/s to the south collides with a 4700 kg truck that is at rest. The car and truck stick together
STatiana [176]

Answer:

Final velocity = 7.677 m/s

KE before crash = 202300 J

KE after crash = 182,702.62 J

Explanation:

We are given;

m1 = 1400 kg

m2 = 4700 kg

u1 = 17 m/s

u2 = 0 m/s

Using formula for inelastic collision, we have;

m1•u1 + m2•u2 = (m1 + m2)v

Where v is final velocity after collision.

Plugging in the relevant values;

(1400 × 17) + (4700 × 0) = (1400 + 1700)v

23800 = 3100v

v = 23800/3100

v = 7.677 m/s

Kinetic energy before crash = ½ × 1400 × 17² = 202300 J

Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J

8 0
3 years ago
A Heavy crate applied a force of 1500 N on a 25m2 piston. What force needs to be applied on a 0.8m2 piston to lift the crate?
Aleks04 [339]
25/1500 is equal to 0.8/x
0.8*1500 is equal to 1200
1200/25 is equal to 48 N
7 0
3 years ago
Read 2 more answers
Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s
Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375  m / s

3 0
3 years ago
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