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jeyben [28]
3 years ago
13

A baseball player hit a ball with an upward velocity of 46

Physics
1 answer:
masya89 [10]3 years ago
3 0
Its height h is feet after t seconds is given by the function h= 16t^2 + 46t + 6.
I hope this helped.
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How much work do you do on a 15 N book in lifting it straight up for a distance of<br> 0.40 meters?
astra-53 [7]

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 :

W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J

So, the magnitude of work done in lifting the book is 6 joules.

7 0
3 years ago
A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the
andreyandreev [35.5K]

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.

d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0
2 years ago
Suppose a rock is dropped off a cliff with an initial speed of 0m/s. What is the rocks speed after 5 secounds, in m/s, if it enc
Tasya [4]

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

3 0
3 years ago
Read 2 more answers
Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at λ = 4.0 × 10-6 C
Len [333]

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

6 0
2 years ago
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p
Kipish [7]

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.

8 0
2 years ago
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