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3 years ago

Ratio of acceleration due to grabity and universal gravitational constant

Physics

1 answer:

creativ13 [48]3 years ago

3 0

Answer:

acceleration due to gravity = 9.8m/s^2

universal gravitational constant= 6.67×10 ^_11 nm^2 kg_2

now, ratio=9.8/6.67×10^_11.

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The sun’s energy is most useful to humans after it is converted to _____.

Trava [24]

The sun’s energy is most useful to humans after it is converted to chemical energy. Living organisms like humans and animals could not directly utilize the energy from the sun because of lack in organs that would allow photosynthesis. Humans and animals obtain the energy from the sun from the intake of food. These foods contain chemical energy which is obtained from the solar energy of the sun. Plants use solar energy for photosynthesis which converts the energy into chemical energy. Then, animals and humans eat the plants obtaining the chemical energy which can readily processed by the body as compared to solar energy.

7 0

3 years ago

Can some answer 7 a and b please

pav-90 [236]

Answer:

a.After $15$ second Mr Comer's speed $=$ $1.7 m/s$

b.Distance travelled by Mr.Comer in $15$ seconds $=$ $18.75\ m$

Explanation:

a. Lets recall our first equation of motion $V_{f} =V_{i} + at$

Now we know that $V_{i} = 0.8m/s$ , $a = 0.06\ m/s^2$ and

$t = 15 \ sec$

Plugging the values we have.

$V_{f} =V_{i} + at$

$V_{f} =0.8 + 0.06 \times 15$

$V_{f} =0.8+ 0.9$

$V_{f} =1.7 m/s$

Then Mr.Comer's speed after $15$ sec $=$ $1.7ms^-1$

Lets find the distance and recall our third equation of motion.

$V_{f} ^2-V{i}^2 = 2as$

So $s =$ distance covered.

Dividing both sides with 2a we have.

$\frac{V_{f} ^2-V{i}^2}{2a} = 2as$

Plugging the values.

$\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as$

$s= 18.75\ m$

So Mr.Comer will travel a distance of $s= 18.75\ m$ .

4 0

3 years ago

When a 25-kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20 below the horizontal, th

Fofino [41]

Answer:

Option E is correct 310N

Explanation:

Given that the force used to push the crate is F = 200N

The force directed 20° below the horizontal

Mass of crate is m = 25kg

Weight of the crate can be determine using

W = mg

g is gravitational constant =9.8m/s²

W = 25×9.8

W = 245 N

Check attachment. For free body diagram and better understanding

Using newton second law along the vertical axis since we want to find the normal force

ΣFy = m•ay

ay = 0, since the body is not moving in the vertical or y direction

N—W—F•Sin20 = 0

N = W+F•Sin20

N = 245+ 200Sin20

N = 245 + 68.4

N = 313.4 N

The normal force is approximately 310 N to the nearest ten

3 0

3 years ago

1. A Zambeef delivery track travels 18 km north, 10 km east, and 16 km south. What is its final displacement from the origin?

VARVARA [1.3K]

Answer:

2km

Explanation:

Given data

We are told that the direction traveled are

North>>>East>>>South

Hence the displacement is defined as the distance away from the initial position is

Initial position =18km

FInal position = 16km

The displacement = 18-16= 2km

Hence the displacement is 2km

6 0

3 years ago

A woman is running around a playground while keeping an eye on her child who is on a swing set. The child is going back and fort

Nutka1998 [239]

Answer:

(a) 1500 m

(b) 2827.43m

Explanation:

Given that the time for one cycle of the swing is 1 s

The radius of the swing, R= 3.0m

The angle covered, \theta, by each swing is a quarter of the circle. i.e.

$\Theta=\frac {\pi}{2}$

Speed of running of women =5 m/s.

Time of running = 5 minutes= 5 x 60 secondes= 300 s.

(a) As distance= (speed) x (time)

So, the required distance= 5 x 300 m= 1500 m.

(b) As there are two swings in one cycle, so, the distance covered in one swing is the length of the circular shown in the figure.

As arc length, $l= \theta R$ , where \theta is the angle, in radian, subtended by the arc at the center, and R is the radius of curvature of the arc.

So, the distance covered by the child in 1 swing $= \frac {\pi}{2}\times 3 m=\frac{3\pi}{2}m$ .

In 1 cycle, there are 2 swings, so distance covered in 1 cycle $= 3 \pi m$ .

Now, in 1 second there is 1 cycle, so in 5 minutes there will be 300 cycles.

So, the total distance covered by the child

$= 3\pi\times 300 m= 2827.43 m$ .

3 0

3 years ago