Answer:
Work done, W = 6 J
Explanation:
It is given that,
Force of gravity acting on the book, weight of the book is 15 N
We need to find the work done in lifting the book straight up for a distance of 0.4 meters.
The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :
So, the magnitude of work done in lifting the book is 6 joules.
Answer:
t = 1.42 s and d = 35.5 m
Explanation:
Given that,
Velocity of a roadrunner is 25 m/s
A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.
We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.
Let d is the distance traveled. So,
d = vt
d = 25 m/s × 1.42 s
d = 35.5 m
Answer:
The rock's speed after 5 seconds is 98 m/s.
Explanation:
A rock is dropped off a cliff.
It had an initial velocity of 0 m/s. And now it is moving downwards under the influence of gravitational force with the gravitational acceleration of 9.8 m/s².
Speed after 5 seconds = V
We know that acceleration = average speed/time
In our case,
g = ((0+V)/2)/5
9.8*5 = V/2
=> V = 2*9.8*5
V = 98 m/s
Answer: 71.93 *10^3 N/C
Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:
∫E*dr=Q inside/εo Q inside is given by: λ*L then,
E*2*π*r*L=λ*L/εo
E= λ/(2*π*εo*r)= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )= 71.93 * 10^3 N/C
Answer:
20.0 cm
Explanation:
Here is the complete question
The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Solution
Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.
Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.
Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m
Now, P' = 1/u + 1/v
1/u = P'- 1/v
1/u = 55.0 D - 1/0.02 m
1/u = 55.0 m⁻¹ - 1/0.02 m
1/u = 55.0 m⁻¹ - 50.0 m⁻¹
1/u = 5.0 m⁻¹
u = 1/5.0 m⁻¹
u = 0.2 m
u = 20 cm
So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.
