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photoshop1234 [79]
3 years ago
8

HELP When the forces are applied in the same direction, how do you determine net force?

Physics
1 answer:
aleksklad [387]3 years ago
5 0

Answer:

Explanation:

If two forces act on an object in the same direction, the net force is equal to the sum of the two forces.

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1. It’s fall and time for the corn maze and bonfire and you just can’t wait. On your way to the farm though a turkey flies out i
Reptile [31]

Acceleration = (change in velocity ( final speed - starting speed))/ (time)

Acceleration = (18-30)/10.5

Acceleration = -12/10.5

Acceleration = -1.14 m/s^2

Distance = 30m/s x 10.5s + 1/2(1.14)(10.5)^2

Distance = 252.2 meters

8 0
3 years ago
A boat heads directly across a river. Its speed relative to the water is 2.6 m/s. It takes it 355 seconds to cross, but it ends
Zigmanuir [339]

Answer:

Vr = 3.24m/s

The boat is going 3.24m/s relative to the bank of the river.

Explanation:

The relative speed of the boat to the bank Vr is the resultant of speed of boat relative to the water Vb and the speed of boat as a result of the water current or wind Vw

Vr = √(Vb^2 + Vw^2) .....1

Given;

Vb = 2.6m/s

Vw = distance downstream/time = 690m/355s

Vw = 1.94m/s

From equation 1 above; substituting the values

Vr = √(2.6^2 + 1.94^2)

Vr = 3.24m/s

The boat is going 3.24m/s relative to the bank of the river.

7 0
3 years ago
Please use this screenshot in any way you can
miskamm [114]
Ball 1 Has uchanging motion because it continually goes up no matter what. Ball 2 has no motion. And ball 3 changing motion because it goes from not moving to moving. Hope this helps
7 0
3 years ago
A microphone is attaced to a spring that is suspended from
kirill115 [55]

To solve this problem we will apply the concept related to the amplitude and the Doppler effect. The difference between the maximum and minimum frequency detected by the microphone would be given by the mathematical function

f_{max} - f_{min} = 2 f_s (\frac{v_0}{v})

Here,

f_s= Source Frequency

v_0 = Speed of microphone

v = Speed sound

f_{max} - f_{min} = 2 f_s (\frac{v_0}{v})

Maximum speed of the microphone is

v_0 = f_{max} - f_{min} * \frac{v}{2f_s}

v_0= \frac{2.1Hz*343m/s}{2*440Hz}

v_0= 0.8185 m/s

Now the amplitude is

A = \frac{v_0}{2\pi/T}

Here T means the Period, then

A= \frac{0.8185 * 2.0}{2\pi}

A= 0.2605m

Therefore the amplitude of the simple harmonic motion is 0.2605m

3 0
4 years ago
An isolated capacitor with capacitance C = 1 µF has a charge Q = 45 µC on its plates.a) What is the energy stored in the capacit
Roman55 [17]

Answer:

a) Energy stored in the capacitor, E = 1.0125 *10^{-3} J

b) Q = 45 µC

c) C' = 1.5 μF

d)  E = 6.75 *10^{-4} J

Explanation:

Capacitance, C = 1 µF

Charge on the plates, Q = 45 µC

a) Energy stored in the capacitor is given by the formula:

E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{2}\\\\E = 1012.5 *10^{-6}\\\\E = 1.0125 *10^{-3} J

b) The charge on the plates of the capacitor will  not change

It will still remains, Q = 45 µC

c)  Electric field is non zero over (1-1/3) = 2/3 of d

From the relation V = Ed,

The voltage has changed by a factor of 2/3

Since the capacitance is given as C = Q/V  

The new capacitance with the conductor in place, C' = (3/2) C

C' = (3/2) * 1μF

C' = 1.5 μF

d) Energy stored in the capacitor with the conductor in place

E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1.5* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{3}\\\\E = 675 *10^{-6}\\\\E = 6.75 *10^{-4} J

4 0
3 years ago
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