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Ymorist [56]
2 years ago
6

Pls help i have test

Physics
2 answers:
Luden [163]2 years ago
8 0

Answer:

A. something pushes or pulls it to stop.

Explanation:

Newtons first law. hope this helps

GenaCL600 [577]2 years ago
4 0
Something pushes or pulls it to stop
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A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. What is the horizontal velocity compo
Sever21 [200]
The horizontal component of the velocity of the ball is calculated by multiplying the speed by the cosine of the given angle. 
                        x-component of speed = (31 m/s)(cos 35°) 
                                                               = 25.39 m/s
Thus, the horizontal velocity component of the ball is 25.39 m/s. 
8 0
3 years ago
Read 2 more answers
HELP!!!!!
devlian [24]

Answer:

c

Explanation:

6 0
3 years ago
Is it possible for a population with a high birth rate to decrease in size.
matrenka [14]
Sure, if the mortality (death) rate is even higher than the birth rate.
7 0
3 years ago
Can anyone help me with this please​
Anastasy [175]

Answer:

1270 J

Explanation:

Recall that the mechanical energy of a system is the addition of the Potential energy and the Kinetic energy at any given time.

As the skier descends, potential energy is converted into kinetic energy, but the total mechanical energy should remain the same.

We see that it is not the case, so that difference is what has gone into thermal energy;  19500 J - 18230 J = 1270 J

7 0
3 years ago
You charge an initially uncharged 89.9-mf capacitor through a 30.5-ω resistor by means of a 9.00-v battery having negligible int
blsea [12.9K]
<span>1) The differential equation that models the RC circuit is :

(d/dt)V_capacitor </span>+ (V_capacitor/RC)​ = (V_source/<span>RC)​​</span>

<span>Where the time constant of the circuit is defined by the product of R*C

Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s


2)
C<span>harge of the capacitor 1.57 time constants

1.57*(2.742) = 4.3048 s

The solution of the differential equation is

</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T)  +  </span>V_capac(∞)

Since the capacitor is initially uncharged V_capac(0) = 0

And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery  V_capac(∞) = 9V 

This means,

V_capac (t) = (-9V)e ^(-t /T)  +  9V

The charge in a capacitor is defined as Q = C*V

Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V = -1.87V +9V

V_capac (4.3048 s) = 7.1275 V

Q (4.3048 s)  = 89.9mF*(7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞)  = 89.9mF*(9V) = 0.8091 C
7 0
3 years ago
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