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mr_godi [17]
3 years ago
12

A bicycle rider traveling east at 10 km/hr sees a blue car pass her appearing to travel west at 50 km/hr. What is the blue car's

apparent speed for an observer riding in a red car traveling 15 km/hr west relative to the Earth? g
Physics
1 answer:
BigorU [14]3 years ago
8 0

Answer:

v_{bR} = 25 km/h towards west

Explanation:

As we know that the speed of the blue car as appear to the bicycle rider is given as

v_{bc} = 50 km/h towards west

also it is given that bicycle is moving at speed of 10 km/h towards East

so here we have

v_{bc} = v_b - v_c

so we have

v_b = v_{bc} + v_c

v_b = -50 + 10 = 40 km/h towards west

now speed of the red car is given as 15 km/h towards west

so here the relative speed of blue car with respect to red car is given as

v_{bR} = v_b - v_R

v_{bR} = 40 - 15 = 25 km/h towards west

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Answer:

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:  

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t  Equation 2

y: The vertical distance the ball moves at time t  

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t  

Known information

We know the following data:

Vi=15 m / s

g =9.8 \frac{m}{s^{2} }

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)  

a) t=1s

y = 15*1 - ½ 9.8*1^{2}=15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

y = 15*1.5 - ½ 9.8*1.5^{2}=22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

y = 15*2 - ½ 9.8*2^{2}= 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

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The mass of the products and reactants are the same on both sides of the equation.

The number of atoms of products and reactants are equal and hence it proves the law of conservation of mass.

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You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo
AlekseyPX

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

(c) When same force is applied to the shopping cart on a wet surface, it will move faster.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

  • F is the applied force
  • Ff is the frictional force

ma = F - Ff

where;

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  • m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net)  = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

Learn more about frictional force here: brainly.com/question/24386803

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