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matrenka [14]
2 years ago
9

Please help me! As quickly as possible

Physics
1 answer:
k0ka [10]2 years ago
7 0

Answer:

1. matter

2. kilograms

3. same

4. gravitational

5. gravity

6. space

7. weightlessness

8. Newton

9. weight

10. more

I HOPE THESE ARE CORRECT AND IT HELPS

You might be interested in
Two charges attract each other with a force of 3.0 n. what will be the force if the distance between them is reduced to one-nint
Schach [20]
The electrostatic force between two charges is given by
F= k_e  \frac{q_1 q_2}{d^2}
where ke is the Coulomb's constant, q1 and q2 are the two charges and d is the distance between the two charges.
We can see from the formula that the force is proportional to \frac{1}{d^2}. This means that if we reduce the distance to one-ninth, i.e. the new distance is
d_{new} =  \frac{1}{9}d
then the force will scale as
 \frac{1}{(d_{new})^2}= \frac{1}{( \frac{1}{9} )^2d^2}= \frac{81}{d^2}
So, the new force will be 81 times stronger than the initial value, therefore the new force is
F_{new}=81 F=81 \cdot 3.0 N=243 N
5 0
3 years ago
What’s the answer anyone??
saul85 [17]

Answer:

5.2L == 5200 mL

3.75 kg == 3750 g

500 mm == 0.5 m

Explanation:

For liters to milliliters, simply multiply by 1000.

For kilograms to grams, simply multiply by 1000

For millimeters to meters, simply divide by 1000.

Cheers.

5 0
3 years ago
Read 2 more answers
An underground cannon launches a cannonball from ground level at a 35 degree angle. the cannonball is shot with an initial veloc
user100 [1]

Answer:

Time = 1.75[s]; Distance traveled = 21.5 [m]; Max height = 15 [m]

Explanation:

First, we have to break down the velocity vector into the X & y components.

(v_{x})_{0} = 15 * cos( 35)= 12.28[m/s]\\(v_{y})_{0} = 15 * sin( 35)= 8.6[m/s]\\\\

To find the time t that lasts the ball of cannon in the air we must use the following equation of kinematics, in this equation the value of y is equal to zero because it will be proposed that the ball lands at the same level that was fired.

y=(v_{y} )_{0}-\frac{1}{2}*g*t^{2}   \\where:\\g=9.81[m/s^2]\\t = time[s]\\y=0[m]

0=8.6*t-\frac{1}{2}*9.81*t^{2}  \\4.905*t^{2}=8.6*t\\ t=1.75[s]

In order to find the distance traveled horizontally from the cannonball, we must use the speed kinematics equation in the X coordinate.

x = (v_{x})_{0}  *t\\x=12.28*1.75\\x=21.5 [m]

In order to find the last value, we must bear in mind that when the cannonball reaches the maximum height, the velocity in the component y is equal to zero, and we can find the value of and with the following kinematic equation

y = (v_{y})_{0} *t+\frac{1}{2} *g*(t)^{2} \\y = 0*t+\frac{1}{2} *9.81*(1.75)^{2}\\ y=15 [m]

6 0
3 years ago
What is the speed of a wave that has a frequency of 2,400 Hz and a wavelength of 0.75
likoan [24]

Answer:

1800 m/s

Explanation:

The equation is v = fλ

λ= 0.75

f = 2400 Hz

V = 2400 × 0.75

V = 1800 m/s

[ you did not give units for wavelength, I assumed it would be m/s]

6 0
3 years ago
46 points :)
IgorC [24]

Answer:

Its is dividing by 2

Explanation:

It starts with 100 them it goes to 50, 25, 12.5 so its a cycle of dividing by 2

3 0
3 years ago
Read 2 more answers
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