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iVinArrow [24]
3 years ago
9

An object began moving at the starting point (0,0) on the graph. The average speed of the object for the entire trip of 5.0 hour

s is:
A. 15mph
B. 17mph
C. 19mph
D. 20.0mph
E. 22mph

Physics
2 answers:
AURORKA [14]3 years ago
8 0
B 17


85 ÷ 5 would be the equation you do to solve
inna [77]3 years ago
4 0

Answer: So in your graph, you can see that after 5 hours, the object is in 85m. So your displacement over the lapse of 5 hours in 85m (because the object starts at 0m), the average speed will be distance over time, so 85m/5h=17m/h

So the correct answer is B, 17 meters per hour.

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Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
Luba_88 [7]

Answer:

V_1=8 V_2

Explanation:

Given that:

  • Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
  • separation distance of capacitor 2, d_2=d
  • separation distance of capacitor 1, d_1=2d
  • quantity of charge on capacitor 2, Q_2=Q
  • quantity of charge on capacitor 1, Q_1=4Q

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

V_1=8 V_2

6 0
3 years ago
The value of a scientific variable ____. A) can change
gregori [183]
The answer is A: can change
5 0
2 years ago
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Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving
Iteru [2.4K]

Explanation:

Let m_p is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.

The magnetic force is balanced by the centripetal force acting on the proton as :

\dfrac{mv^2}{r}=qvB

r is the radius of path,

r=\dfrac{mv}{qB}

Time period is given by :

T=\dfrac{2\pi r}{v}

T=\dfrac{2\pi m_p}{qB}

Frequency of proton is given by :

f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}

The wavelength of radiation is given by :

\lambda=\dfrac{c}{f}

\lambda=\dfrac{2\pi m_pc}{qB}

So, the wavelength of radiation produced by a proton is \dfrac{2\pi m_pc}{qB}. Hence, this is the required solution.

3 0
3 years ago
Will mark brainliest!
MaRussiya [10]

Answer:

Explanation:

Fa - u*m*g = m*a

Fa = u*m*g + m*a

Fa - m*a = u*m*g

u = \frac{Fa - m*a}{m*g}

8 0
3 years ago
Does gravity have an effect on the potential difference of a battery?
Sonja [21]

Answer: The correct answer is "No".

Explanation:

Gravity: It is the force which causes object to fall on the earth. It is the force which attracts bodies towards each other.

Potential difference: It is defined as the potential acting between the two points. The work done in moving the unit positive charge from one location to the another location.

The potential difference in battery is caused by the electrodes. There are two terminals in battery: Negative terminal which is at lower potential and Positive terminal which is at higher potential. It forces the electrons to flow in the circuit which constitutes the current.

The gravity and the potential difference have no relation between them.

Therefore, gravity have no effect on the potential difference of a battery.

5 0
3 years ago
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