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MakcuM [25]
3 years ago
14

A bus travels north on some busy city streets for 2.5 km, and a trip

Physics
1 answer:
d1i1m1o1n [39]3 years ago
8 0

Answer:

V = 4.63 m/s

V = 11.31 m/s

Explanation:

Given,

The distance traveled by the bus, towards north, d = 2.5 km

                                                                                     = 2500 m

The time taken by the trip is, t = 9 min

                                                  = 540 s

The velocity of the bus,

                                       V = d / t

                                           = 2500 / 540

                                          = 4.63 m/s

At another  point, the bus travels at a constant speed of v = 18 m/s

Therefore the velocity becomes

                                                V = (4.63 + 18)/2

                                                   = 11.31 m/s

Hence, the velocity of the  bus, V = 11.31 m/s

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A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the
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Answer:

Option B. 2.8 s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 27 m/s

Angle of projection (θ) = 30

Acceleration due to gravity (g) = 9.8 m/s²

Time of flight (T) =?

The time of flight of the ball can be obtained as follow:

T = 2uSineθ / g

T = 2 × 27 × Sine 30 / 9.8

T = 2 × 27 × 0.5 / 9.8

T = 27 / 9.8

T = 2.8 s

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7 0
3 years ago
An electric field of intensity 3.7 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.3
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Answer:

a)906.5 Nm^2/C

b) 0

c) 742.56132 N•m^2/C

Explanation:

a) The plane is parallel to the yz-plane.

We know that

flux ∅= EAcosθ

3.7×1000×0.350×0.700=906.5 N•m^2/C

(b) The plane is parallel to the xy-plane.

here theta = 90 degree

therefore,

0  N•m^2/C

(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.

therefore, applying the flux formula we get

3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C

4 0
3 years ago
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3 years ago
a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b
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Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

In an attempt o move a crate;

Force applied = 2470 N

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We know that the work done is defined as the force used to move an object to a distance.

Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.

Work done is defined as:

Work = Force*distance covered in the direction of the force

3650 = 2470*distance

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distance = 1.48 meters

4 0
3 years ago
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