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HACTEHA [7]
3 years ago
12

What is the average speed of a car that travelled 400 miles in 6 hours?

Physics
2 answers:
Reil [10]3 years ago
8 0

Answer:

About 66 miles per hour

Explanation:

Based on the information given we can assume the car traveled the same number of miles every hour meaning all we need to do is divide.

400/6 ≈ 66 miles per hour

Ahat [919]3 years ago
7 0

Answer

66 mph

Explanation:

400 ÷ 6 = 66.666666666666666.......

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A plane starting at rest at one end of a runway undergoes a uniform acceleration of 4.8 m/s
yKpoI14uk [10]
  • initial velocity=0m/s=u
  • Acceleration=a=4.8m/s^2
  • Time=t=15s

Final velocity be v

\\ \sf\longmapsto v=u+at

\\ \sf\longmapsto v=0+4.8(15)

\\ \sf\longmapsto v=72m/s

8 0
2 years ago
A horizontal desk surface measures 1.7 m by 1.0 m. If the Earth's magnetic field has magnitude 0.42 mT and is directed 68° below
AlexFokin [52]

Answer:

The magnetic flux through the desk surface is 6.6\times10^{-4}\ T-m^2.

Explanation:

Given that,

Magnetic field B = 0.42 T

Angle =68°

We need to calculate the magnetic flux

\phi=BA\costheta

Where, B = magnetic field

A = area

Put the value into the formula

\phi=0.42\times10^{-3}\times1.7\times1.0\cos22^{\circ}

\phi=0.42\times10^{-3}\times1.7\times1.0\times0.927

\phi=6.6\times10^{-4}\ T-m^2

Hence, The magnetic flux through the desk surface is 6.6\times10^{-4}\ T-m^2.

3 0
3 years ago
once you decide that it is safe to move out of traffic, what steps should you follow to ensure safe passage?
Solnce55 [7]

First put your turn signal on, next check for any ongoing traffic and wait until it is clear lastly start to drift into the lane you need to clear away from traffic

4 0
3 years ago
If an electronin an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T
Anni [7]

Answer:

Explanation:

Given that,

Force is downward I.e negative y-axis

F = -2 × 10^-14 •j N

Magnetic field is westward, +x direction

B = 8.3 × 10^-2 •i T

Charge of an electron

q = 1.6 × 10^-19C

Velocity and it direction?

Force in a magnetic field is given as

F = q(V×B)

Angle between V and B is 270, check attachment

The cross product of velocity and magnetic field

F =qVB•Sin270

2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2

Then,

v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)

v = 1.51 × 10^6 m/s

Direction of the force

Let x be the direction of v

-F•j = v•x × B•i

From cross product

We know that

i×j = k, j×i = -k

j×k =i, k×j = -i

k×i = j, i×k = -j OR -k×i = -j

Comparing -k×i = -j to given problem

We notice that

-F•j = q ( -V•k × B×i)

So, the direction of V is negative z- direction

V = -1.51 × 10^6 •k m/s

6 0
2 years ago
I don’t understand I need help asap
hammer [34]
For the first question, you got them right, for the two you left blank, initial(beginning) velocity: 2 m/s the final velocity is: 12 m/s
3 0
3 years ago
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