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Murrr4er [49]
3 years ago
12

NEED THIS DONE ASAP HELP

Physics
1 answer:
ExtremeBDS [4]3 years ago
4 0

Answer: Exercise Physiology

Explanation:

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A small, positively charged ball is moved close to a large, positively charged ball. which describes how the small ball likely r
NNADVOKAT [17]

Answer;

-it will move away from the large ball because like charges repel.

Explanation;

-Electric force is the force that pushes apart two like charges, or that pulls together two unlike charges. The basic law of electrostatics Like charges of electricity repel each other, whereas unlike charges attract each other.

When small, positively charged ball is moved close to a large, positively charged ball it would be pushed away from the large positively charged ball since they are both positively charged. One has to put in energy to try to move the small ball closer to the large ball. The closer one try to move it to the large ball, the more energy one has to put in, so the more electrical potential energy the small ball would have.

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Is blacklight the same as ultraviolet?
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Blacklight is made of a type of ultraviolet light called UV-A, but blacklight is not the same as ultraviolet light. Blacklight lamps only emit in the lower energy range of ultraviolet light.
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A dog leaps horizontally off a 70 m cliff with a speed of 6 m/s, how far from the base will the dog land?
iris [78.8K]

Answer:

\Delta x=22.67786838m

Explanation:

Let's use projectile motion equations. First of all we need to find the travel time. So we are going to use the next equation:

y-y_0=v_o*sin(\theta)*t-\frac{1}{2}*t^2 (1)

Where:

y=Final\hspace{3}position\hspace{3}at\hspace{3}y-axis

y_o=Initial\hspace{3}position\hspace{3}at\hspace{3}y-axis

v_o=initial\hspace{3}velocity

t=travel\hspace{3}time

g=gravity\hspace{3}constant

\theta=Initial\hspace{3}launch\hspace{3}angle

In this case:

\theta=0

Because the dog jumps horizontally

Let's asume the gravity constant as:

g=9.8

y=0

Because when the dog reach the base the height is 0

y_o=70

v_o=6

Now let's replace the data in (1)

y_o-\frac{1}{2} *(9.8)*t^2+70

Isolating t:

t=\pm\sqrt{\frac{2*70}{9.8} } =3.77964473

Finally let's find the horizontal displacement using this equation:

\Delta x=v_o*cos(\theta)*t

Replacing the data:

\Delta x=6*1*3.77964473=22.67786838m

8 0
3 years ago
If the volume of a container of gas remains constant, what will happen to the pressure of a gas if you increase the temperature?
Scorpion4ik [409]

Answer:

This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure.

Explanation:

Hope it helps

4 0
2 years ago
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