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oksano4ka [1.4K]
3 years ago
9

A comet is cruising through the Solar System at a speed of 50,000 kilometers per hour for 4.5 hours time. What is the total dist

ance traveled by the comet during this time?
Physics
1 answer:
PSYCHO15rus [73]3 years ago
4 0

Answer:

The total distance traveled by the comet is 225000 [km]

Explanation:

In order to determine the travel distance we must use the following kinematics equation:

x = v*t

where:

x = distance traveled [km]

v = velocity = 50000 [km/h]

t = time = 4.5 [h]

x = 50000*4.5 = 225000 [km]

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Which factors affect the gravitational force between objects? Check all that apply. difference in speed of objects direction of
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The surface temperature of a star can be estimated based on the star’s a. size. c. age. b. color. d. mass. Please select the bes
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IP A spark plug in a car has electrodes separated by a gap of 6.5×10−2 in . To create a spark and ignite the air-fuel mixture in
emmainna [20.7K]

Answer:

a.) V=5283.2 \ V

b.) V=4064\ V

Explanation:

<u>Electric Field and Potential Difference</u>

There are several conditions that must be met for a spark to be created into an air gap. Once the physical conditions are fixed, a minimum electric field is necessary for the spark to be initiated. Let s be the separation between the electrodes and V their potential difference. The electric field is

a.)

\displaystyle E=\frac{V}{s}

Solving for V

V=E.s

The separation is

s= 5.5\cdot 10^{-2}\ in=0.001651 \ m

Thus the potential difference is

V=3.2\cdot 10^{6}\ V/m\times 0.001651 \ m

V=5283.2 \ V

b.) If the separation was greater, the applied voltage needs to be greater if the electric field has to be constant. One possible measure to keep electrodes as close as possible is to build them as sharp edges. It gives the spark an easier path to travel to.

If the separation is 0.05 inches =0.00127 m

V=3.2\cdot 10^{6}\ V/m\times 0.00127 \ m

V=4064\ V

7 0
3 years ago
a charge of 7.2x10^-5C is placed in an Electric field with a strength of 4.8x10^5N/C. if the Electric potential energy of the ch
Alecsey [184]

Answer:

d= 2.2 m

Explanation:

The data given in the question is

Charge : q =7.2 \times 10^{-5} C

Electric field Strength : E = 4.8 \times 10^{5} N/C

Electrical potential Energy : U = 75J

Let, the distance be "d"

So, the formula for Electrical potential energy is

U= q \times E \times d

Simplified formula for distance become

d=\frac{U}{q \times E}

Now , insert the value

d=\frac{75 J}{7.2 \times 10^{-5}C \times 4.8 \times 10^5 N/C}

or,

d=2.17 m

Rounding off

d= 2.2 m

3 0
3 years ago
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