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Yuri [45]
3 years ago
11

Two ice skaters are initially at rest. They push off and move in opposite directions. Ice Skater 1 has a mass of 90 kg and a vel

ocity of -10 meters per second. Ice Skater 2 has a mass of 105 kg. What is his velocity?
Physics
1 answer:
makvit [3.9K]3 years ago
4 0

Answer:

vB = 8.57[m/s]

Explanation:

Since there is no external force that generates momentum, the amount of momentum is conserved so you will have the following equation

m_{A}*v_{A}=m_{B}*v_{B}\\Where:\\m_{A}= 90[kg]\\v_{A}=10[m/s]\\m_{B}=105[kg]\\\\Therefore:\\\\v_{B}=\frac{m_{A}*v_{A}}{m_{B}}\\ v_{B}=\frac{90*10}{105}\\v_{B}=8.57[m/s]

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A car travels from 20-meters to 60-meters in 10 seconds. Calculate the car's speed.
slamgirl [31]

Answer:

8 m/s

Explanation:

All you have to do here is add 20 and 60 (giving you 80) and dividing by 10 seconds. 80/10= 8 m/s

3 0
3 years ago
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Alecsey [184]
I think it's A Tsunami. Or an Earthquake

6 0
2 years ago
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Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0
3 years ago
Listen →
Sedaia [141]

Answer:

V = 20.5 m/s

Explanation:

Given,

The mass of the cart, m = 6 Kg

The initial speed of the cart, u = 4 m/s

The acceleration of the cart, a = 0.5 m/s²

The time interval of the cart, t = 30 s

The final velocity of the cart is given by the first equation of motion

                              v = u + at

                                  = 4 + (0.5 x 30)

                                = 19 m/s

Hence the final velocity of cart at 30 seconds is, v = 19 m/s

The speed of the cart at the end of  3 seconds

                                    V = 19 + (0.5 x 3)

                                       = 20.5 m/s

Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s

6 0
3 years ago
What is the temperature of a 3.72 mm cube (e=0.288) that radiates 56.6 W?
blsea [12.9K]

Answer:

The temperature is 2541.799 K

Explanation:

The formula for black body radiation is given by the relation;

Q = eσAT⁴

Where:

Q = Rate of heat transfer 56.6

σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)

A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²

e = emissivity = 0.288

T = Temperature

Therefore, we have;

T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴

T  =  2541.799 K

The temperature = 2541.799 K.

7 0
3 years ago
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