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Pepsi [2]
3 years ago
10

A hot–air balloon is moving at a speed of 10.0 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–d

irection at a velocity of +4 meters/second with respect to himself. If the ball lands after 30 seconds, about how far does it land horizontally from the point at which it was released?
A. 120 meters B. 125 meters C. 300 meters D. 420 meters
Physics
1 answer:
yulyashka [42]3 years ago
3 0

Option (D) is correct.

The balloon lands horizontally at a distance of 420 m from a point where it as released.

Explanation:

Velocity of air balloon along +X axis =10 m/s

velocity of ball=4 m/s along + X axis

the velocity of balloon gets added to the velocity of ball. So the resultant velocity of the balloon=10+4 = 14 m/s

time taken= 30 s

The distance traveled is given by d= v t

d= 14 (30)

d= 420 m

Thus the balloon lands horizontally at a distance of 420 m from a point where it as released.

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nirvana33 [79]
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Hello,help me with this out please i need it hurry but please ensure your answer is correct..I attach here with my question.
lilavasa [31]

Answer:

.409 N

Explanation:

For this to balance, the moments around the fulcrum must sum to zero.

On the left you have   .21   ( is that down? I will assume it is)

      Counterclockwise moments :

        .21 * 40     +  1.0 * 20    

     Clockwise moments :

        .5 * 20     +     F * 45

these moments must equal each other

.21*40 + 1 *20   =  .5 * 20 + F * 45

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7 0
2 years ago
Two cars leave towns 680 kilometers apart at the same time and travel toward each other. one car's rate is 10 kilometers per hou
Evgen [1.6K]

D = distance between the cars at the start of time = 680 km

v₁ = speed of one car

v₂ = speed of other car = v₁ - 10

t = time taken to meet = 4 h

distance traveled by one car in time "t" + distance traveled by other car in time "t" = D

v₁ t + v₂ t = D

(v₁ + v₂) t = D

inserting the values

(v₁ + v₁ - 10) (4) = 680

v₁ = 90 km/h

rate of slower car is given as

v₂ = v₁ - 10

v₂ = 90 - 10 = 80 km/h

5 0
3 years ago
2) Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another
worty [1.4K]

Answer:

m_1 / m_2 = sqrt (1 / 2)

Explanation:

Given:

- Initial velocity of both skaters V_i = 0

- Velocity of skater 1 after push = V_1

- Velocity of skater  after push = V_2

- Distance traveled by skater 1 = s_1

- Distance traveled by skater 2 = s_2

- s_1 = 2*s_2

- Accelerations of both skaters to halt is equal

Find:

What is the ratio m1/m2 of their masses

Solution:

- Apply conservation of momentum for two skaters just before and after the push as follows:

                                              P_i = P_f

                                  0 = m_1*V_1 - m_2*V_2

- Evaluate:                 m_1 / m_2 = ( V_2 / V_1 )

- Apply Conservation of Energy on both skaters as follows:

- Skater 1:

                               0.5*m_1*V_1^2 = u_k*m_1*g*s_1

-Simplify:                      0.5*V_1^2 = u_k*g*(2*s_2)

- Skater 2:

                               0.5*m_2*V_2^2 = u_k*m_2*g*s_2

-Simplify:                      0.5*V_2^2 = u_k*g*s_2

- Divide the two energy equations for skaters:

                                    (V_1 / V_2)^2 = 2

                                    (V_2 / V_1)^2 = 1 / 2

- simplify:                     (V_2 / V_1) = sqrt (1 / 2)

-Hence from earlier momentum conservation results:

                                  m_1 / m_2 = ( V_2 / V_1 ) = sqrt (1 / 2)

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Answer:

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