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Oxana [17]
4 years ago
14

Question 1 (1 point)

Physics
1 answer:
puteri [66]4 years ago
3 0

Answer:

250 N

Explanation:

a =  \frac{vf - vi}{t}

Vf=final velocity

Vi =initial velocity

=70-20/2=25m/s^2

F=ma

=10kg * 25m/s^2

=250N

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A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ
valentinak56 [21]

Explanation:

The given data is as follows.

                    mass = 0.20 kg

              displacement = 2.6 cm

              Kinetic energy = 1.4 J

       Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

         Total energy = Kinetic energy + spring potential energy

                           = 1.4 J + 2.2 J

                            = 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So,      K.E = Total energy

                = 3.6 J

Also, we know that

                  K.E = \frac{1}{2}mv^{2}_{m}

or,                   v = \sqrt{\frac{2K.E}{m}}

                        = \sqrt{2 \times 3.6 J}{0.2 kg}

                        = \sqrt{36}

                        = 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0
3 years ago
On a straight road (taken to be in the x direction) you drive for an hour at 60 km per hour, then quickly speed up to 120 km per
Luda [366]

Answer:

The average velocity is 180 km/hr

Explanation:

Given;

initial velocity, u = 60 km per hour

final velocity, v = 120 km per hour

initial time = 1 hour

final time = 2 hour

Initial position = 60 km/h x 1 hour = 60 km

final position = 120 km/h x 2 hour = 240 km

The average velocity is given by;

V_{avg} = \frac{Final \ position\  - \ Initial \ position}{final \ time\  - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\  - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr

Therefore, the average velocity is 180 km/hr

3 0
3 years ago
Changing the wavelength of a light wave also changes its:_____.
Art [367]

Answer:

frequency

Explanation:

speed of light = <u>constan</u>t = wavelength * frequency

                                                changing WL or F will change the other

7 0
2 years ago
A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 5 m from the bulb,
beks73 [17]

Answer:

a) 0.477 W/m²

b) 13.407 N/C

c) 18.96 N/C

Explanation:

P = Power = 150 W

r = Distance = 5 m

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

a) Average intensity

\bar{S}=\frac{P}{A}\\\Rightarrow \bar{S}=\frac{150}{4\pi r^2}\\\Rightarrow \bar{S}=\frac{150}{4\pi 5^2}\\\Rightarrow \bar{S}=0.477\ W/m^2

∴ Average intensity is 0.477 W/m²

b) Rms value

\bar{S}=c\epsilon_0E_{rms}^2\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{S}}{c\epsilon_0}}\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{0.477}}{3\times 10^8\times 8.854\times 10^{-12}}}\\\Rightarrow E_{rms}=13.407\ N/C

∴ Rms value of the electric field is 13.407 N/C

c) Peak value

E_0=\sqrt 2E_{rms}\\\Rightarrow E_0=\sqrt 2\times 13.407\\\Rightarrow E_0=18.96\ N/C

∴ Peak value of the electric field is 18.96 N/C

8 0
3 years ago
A negative charge -0.550 HC exerts an upward 0.700 N force on an unknown charge 0.220 m directly below it. (a) What is the unkno
Sonja [21]

Answer:

Part a)

q_2 = -6.8 \mu C

Part b)

F = 0.700 N

direction = downwards

Explanation:

As we know that the negative charge will experience the force due to some other charge below it

the force is given as

F = 0.700 N

now we know that

F = \frac{kq_1q_2}{r^2}

now plug in all data

0.700 = \frac{(9 \times 10^9)(0.550\mu C)q_2}{0.220^2}

0.700 = 1.022\times 10^5 q_2

q_2 = -6.8 \mu C

since this is a repulsion force so it must be a negative charge

Part b)

As per Newton's III law it will exert equal and opposite force on it

So here the force on the charge below it will be same in magnitude but opposite in direction

so here we have

F = 0.700 N

direction = downwards

4 0
3 years ago
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