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Helga [31]
3 years ago
11

10.A car is travelling at a constant speed of 27m/s. The driver looks away from the road for a 2.0s to tune in a station on the

radio. How far does the car go during this time?
Physics
1 answer:
Korvikt [17]3 years ago
5 0

Explanation:

Distance = speed × time

d = (27 m/s) (2.0 s)

d = 54 m

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James took two pea plants, placing one in a dark closet and the other on a sunny window sill. Both are located in air-conditione
Fudgin [204]

The constant is the temperature of the air that the plants get.

The independent variable is the thing that YOU control.  That's the amount of sunlight each plant gets.

The <em>dependent variable</em> is anything that's caused by changes in the independent variable.  That's the growth of the plants.

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SashulF [63]
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7 0
3 years ago
A 180 lb crate is on the ground, and a strong rope is attached. You need to move it across the basement floor, which has a coeff
jekas [21]

Answer:

F = 505.13 N

Yes it is better to pull the rope rather than push it

Explanation:

Let the force is applied at an angle of 60 degree

so we will have net vertical force on the crate is given as

F_n + Fsin60 = mg

here we know

m = 180 lb

m = 81.65 kg

F_n = 81.65(9.81) - Fsin60

F_n = 801 - 0.866 F

now friction force on the crate is given as

F_x = \mu F_n

Fcos60 = 0.7(801 - 0.866 F)

0.5F + 0.61F = 560.7

F = 505.13 N

Yes it is better to pull the rope rather than push it

6 0
3 years ago
A small plastic ball with a mass of 7.00 10-3 kg and with a charge of +0.155 µC is suspended from an insulating thread and hangs
adell [148]

Answer:

the magnitude of the charge Q on each plate is 3.053 *10^{-8} \ C

Explanation:

Given that :

mass (m) = 7.00 *10 ^{-3} \ kg

charge (q) = +0.155 µC = +0.155 *10^{-6}\ C

angle \theta = 30^0 \ C

Area A on each plate = 0.0135 m²

From the diagram below;

tan \ \theta = \frac{Eq}{mg}    ----- equation (1)

Also by using Gauss Law ;

Q = \epsilon_0 \phi

Q = \epsilon_0EA     ----- equation (2)

Combination equation 1 and 2 together ; we have

Q = \frac{\epsilon_0\ * \ m\ *\ g \ \ * \ tan \theta \ * \ A}{q}

Q = \frac{(8.85*10^{-12}C^2/N.m^2 )\ * \ (7.00*10^{-3} kg)\ *\ (9.8 m/s^2) \ \ * \ tan \(30 \ * \ (0.0135 m^2)}{0.155*10^{-6}\ C}

Q = 3.053 *10^{-8} \ C

8 0
3 years ago
Using the scientific definition of work, does moving an object a greater amount of distance always require a greater amount of w
tester [92]
The answer is no. If you are dealing with a conservative force and the object begins and ends at the same potential then the work is zero, regardless of the distance travelled. This can be shown using the work-energy theorem which states that the work done by a force is equal to the change in kinetic energy of the object.
W=KEf−KEi
An example of this would be a mass moving on a frictionless curved track under the force of gravity.
The work done by the force of gravity in moving the objects in both case A and B is the same (=0, since the object begins and ends with zero velocity) but the object travels a much greater distance in case B, even though the force is constant in both cases.

3 0
4 years ago
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