1. A horizontal force of 50 N is applied to push a desk 40 m across a

The four balls have different masses and speeds.

g100num [7]

Answer:

C, A, D, B

Explanation:

You just need to calculate the momentum of each ball, by p=m*v

pA = 0.7kg*15m/s = 10.5 kg-m/s

pB = 1.2kg*5.5m/s = 6.6 kg-m/s

pC = 2.5kg*5m/s = 12.5 kg-m/s

pD 5kg*1.5 = 7.5 kg-m/s

Rankin them, we will have: C, A, D, B.

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3 years ago

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What is the voltage drop across the 30 q resistor? A. 120 v B. 30 V c. 2 v D. 60 V

ryzh [129]

Answer:

120v 120v 120v 120v 120v

4 0

3 years ago

A 1200-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 9,000-kg truc

UNO [17]

Answer

given,

mass of car (m) = 1200 Kg

speed of cur, u = 25 m/s

mass of truck(M) = 9000 Kg

speed of truck, u' = 20 m/s

v = 18 m/s

a) using conservation of momentum

m u + M u' = m v + M v'

1200 x 25 + 9000 x 20 = 1200 x 18 + 9000 x v'

9000 v' = 188400

v' = 20.93 m/s

b) To calculate losses, we will find the kinetic energies before & after collision. Any difference would give us the losses (in energy form).

(K E)₁= (K E)₂+ Losses

losses = (K E)₂ - (K E)₁

= $\dfrac{1}{2}(mv^2 + Mv'^2)- \dfrac{1}{2}(mu^2 + Mu'^2)$

= $\dfrac{1}{2}(1200 \times18^2 +9000 \times 20.93^2)- \dfrac{1}{2}(1200 \times 25^2 +9000\times 20^2)$

=9038 J

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3 years ago

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What variables affect density weight,conductivity,color,volume, or mass

tatiyna

Answer:

Density is affected by volume and mass.

Explanation:

Density is defined as the quantity of mass per unit of volume, or expressed mathematically, d = m/v.

6 0

3 years ago

Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on

nexus9112 [7]

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂ -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C (OR) q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC (OR) q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

8 0

3 years ago