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andrew-mc [135]
3 years ago
10

Theweight ofof body is 420N .Calculate its mass​

Physics
1 answer:
Nady [450]3 years ago
6 0

Answer:

42.87

Explanation:

Google

<u>OR</u>

<u></u>

420/9.8 which equals 42.87.

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In the great shopping cart race, two students push on shopping carts. A having twice the mass of B, with the same force applied
Lesechka [4]

K = 1/2 m x v^2

m = mass on the cart

V = velocity imparted to the cart

KA = 1/2 mA x vA^2.......................(1)

KB = 1/2 mB x vB^2........................(2)

Diving equation 1 by equation 2, we get -

KA/KB = mA/mB

= 2

KA = 2 x KB

Option A is correct

6 0
3 years ago
Series circuits are characterized by the fact that there is a single pathway by which charge can travel. True or false?
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Hope this helps
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Zane
6 0
3 years ago
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The earth____ around the sun because the earth___ we have night and day
Harrizon [31]

Answer:

The Earth Revolves around the sun because the Earth Tilts<u>,</u> we have night and day.

Explanation:

Hope this helps! <3

4 0
3 years ago
A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

7 0
3 years ago
A wave has a wave length of 20mm and a frequency of 5 hertz. What is the speed?
zmey [24]

Speed = (wavelength) x (frequency

Speed = (.020 m) x (5 / sec)

Speed = 0.1 m/s
7 0
3 years ago
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