Answer:
15.23.....................

Answer:
I think it would be 15.23 not so sure but

hope this helps! (:

hope this helps! (:

A construction foreman exerts 1300 Newtons of force trying to move a 1200-kg block of concrete. How many Joules of work does he perform?

Air trapped in the cooling system could create undesirable areas of combustion heat buildup in the engine called _____.

An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equal amounts of energy are added to them. Assuming no melting or vaporization occurs, which of the following can be concluded about the final temperature TA of substance A and the final temperature TB of substance B?a) TA > TBb) TA < TBc) TA = TBd) More information is needed

Newton's law of gravity was inconsistent with Einstein's special relativity becausethe distance, and thus the force, depends on the reference frame chosentime dilation slows the apparent acceleration, reducing the forcemass, and thus the force, depends on the reference frame chosenNewton didn't know about Einstein

Air trapped in the cooling system could create undesirable areas of combustion heat buildup in the engine called _____.

An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equal amounts of energy are added to them. Assuming no melting or vaporization occurs, which of the following can be concluded about the final temperature TA of substance A and the final temperature TB of substance B?a) TA > TBb) TA < TBc) TA = TBd) More information is needed

Newton's law of gravity was inconsistent with Einstein's special relativity becausethe distance, and thus the force, depends on the reference frame chosentime dilation slows the apparent acceleration, reducing the forcemass, and thus the force, depends on the reference frame chosenNewton didn't know about Einstein

**Answer:**

**3.06 seconds time passes before the watermelon has the same velocity **

**watermelon going at speed 59.9 m/s**

**watermelon traveling when it hits the ground at speed is 79.19 m/s**

**Explanation:**

given data

height = 320 m

speed = 30 m/s

to find out

How much time passes before the watermelon has the same velocity and How fast is the watermelon going and How fast is the watermelon traveling

solution

we will use here equation of motion that is

v = u + at ....................1

here v is velocity 30 m/s and u is initial speed i.e zero and a is acceleration i.e 9.8 m/s²

put the value and find time t

30 = 0 + 9.8 (t)

t = 3.06 s

**so 3.06 seconds time passes before the watermelon has the same velocity **

and

we know superman cover distance is = velocity × time

so distance = 30 × t

and distance formula for watermelon is

distance = ut + 0.5×a×t² .............2

here u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and h is 30 × t

30 × t = 0 + 0.5×9.8×t²

t = 6.12 s

so by equation 1

v = u + at

v = 0 + 9.8 ( 6.12)

v = 59.9 m/s

**so watermelon going at speed 59.9 m/s**

and

watermelon traveling speed formula is by equation of motion

v² - u² = 2as ......................3

here v is speed and u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and s is distance i.e 320 m

v² - 0 = 2(9.8) 320

v = 79.19 m/s

**so watermelon traveling when it hits the ground at speed is 79.19 m/s**

maximum speed of cheetah is

speed of gazelle is given as

Now the relative speed of Cheetah with respect to Gazelle

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

so by rearranging the terms we can say

so above is the relation between all given variable

**Answer:**

Radio waves

**Explanation:**

Radio wavs are electromagnetic waves.

Hope this helped!

The answer is Radio Waves because it is electromagnetic

**Answer:**

The image formed by a convex mirror will always have its smaller than the size of the object no matter what the position of the object.

**Explanation:**

The image formed by a convex mirror will always have its smaller than the size of the object no matter what the position of the object.

Also notice that convex mirror always makes virtual images.

Another feature of the convex mirror is that an upright image is always formed by the convex mirror.

An important mirror formula to remember which is applicable for both convex and mirrors

- 1/f= 1/u + 1/v

Here:

'u' is an object which gets placed in front of a spherical mirror of focal

length 'f' and image 'u' is formed by the mirror.

**Answer:**

right side up

**Explanation:**

B) d/√2

C) d/4

D) 2d

E) d/2

**Answer:b**

**Explanation:**

Given

Force of attraction is F when charges are d distance apart.

Electrostatic force is given by

where k=constant

and are charges

d=distance between them

In order to double the force i.e. 2F

divide 1 and 2 we get

b. Determine the value of the coefficient of friction between the tires and the road.

**(a)** It's the force of (static) friction that keeps the car on the road and prevents it from skidding, and this friction is directed toward the center of the curve.

Recall that centripetal acceleration has a magnitude *a* of

*a* = *v* ² / *R*

where

*v* = tangential speed

*R* = radius of the curve

so that

*a* = (35 m/s)² / (215 m) ≈ 5.69767 m/s² ≈ 5.7 m/s²

Parallel to the road, the only force acting on the car is friction. So by Newton's second law, we have

∑ *F* = *Fs* = *m**a*

where

*Fs* = magnitude of static friction

*m* = mass of the car

Then

*Fs* = (950 kg) (5.7 m/s²) ≈ 5412.79 N ≈ **5400 N**

**(b)** Perpendicular to the road, the car is in equilbrium, so its weight and the normal force of the road on the car are equal in magnitude. By Newton's second law,

*N* - *W* = 0

where

*N* = magnitude of normal force

*W* = weight

so that

*N* = *W* = *m g* = (950 kg) (9.8 m/s²) = 9310 N

Friction is proportional to the normal force by a factor of *µ*, the coefficient of static friction:

*Fs* = *µ**N*

Assuming 35 m/s is the maximum speed the car can travel without skidding, we find

*µ* = *Fs* / *N* = (5400 N) / (9310 N) ≈ 0.581395 ≈ **0.58**