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3 years ago

Please help me answer

Physics

2 answers:

STatiana [176]3 years ago

4 0

D. 55 Kg mass moving at 1 m/s, is the correct answer.

The more mass the object has, the greater inertia is. It doesn't matter how fast the object is going because inertia is depended on mass.

yulyashka [42]3 years ago

3 0

I don’t know why it want let me say it so just comment on it !!

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Which is the quickest way to accurately take your heart rate

Thepotemich [5.8K]

I would say by putting two fingers under your chin or putting two fingers on the back of your wrist, hope i helped ! :)

3 0

3 years ago

A capacitor is charged until its stored energy is 7.54 J. A second capacitor is then connected to it in parallel. If the charge

Ivan

Answer:

2 J

Explanation:

A charged capacitor of capacitance $C_1$ with energy of 7.54 J, is connected in parallel with another capacitor $C_2$ , so the charge is equally distributed between them.

(a) The energy stored in the capacitor before it being connected to the other capacitor is:

$U_O=q_0^2/2C_1=7.54 J\\$

The energy stored in the electric field is the sum of the energies of the two capacitors:

$U=U_1+U_2\\U=q_1^2/2C_1+q_2^2/2C_2$

since the charge equally distributed, $q_1$ = $q_2$ = $q_o/2$ . and since they are connected in parallel the potential difference on both of them is the same :

$V_1=V_2\\q_1/C_1=q_2/C_2\\q_0/C_1=q_0/C_2\\C_1=C_2=C_3\\$

hence,

$U=q_0^2/8C+q_0^2/8C\\U=q_0^2/4C\\U=2J$

8 0

3 years ago

ASAP

scoundrel [369]

Answer:

Explanation:

Hooke's law! F(spring)=-kx

There's no tricky square law here. The spring constant doesn't change, only x (distance stretched) changes. Therefore, if distance is halved, Force will be halved.

5 0

3 years ago

A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$