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A light year is defined as the distance that light can travel in 1 year. What is the value of 1 light year in meter? Show your c

Ipatiy [6.2K]

Answer:

d=9.462×10^15 meters

Explanation:

Relation between distance, temps and velocity:

d=v*t

t=1year*(365days/1year)*/(24hours/1day)*(3600s/1h)=31536000s

So:

1 light year=d=3*10^8m/s*3.154*10^7s=9.462×10^15 meters

6 0

3 years ago

How long must a tow truck apply a force of 600 N to increase the speed of a 1,500 kg car at rest to 2 m/s?

slega [8]

The time the truck must apply the given force to increase its speed to given value is 5 s.

The given parameters;

applied force, F = 600 N
mass of the truck, m = 1,500 kg
speed of the truck, v = 2 m/s

The force applied to the truck is determined by Newton's second law of motion; which states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.

F = ma

$F = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{1500 \times 2}{600} \\\\t = 5 \ s$

Thus, the time the truck must apply the given force to increase its speed to given value is 5 s.

Learn more here:brainly.com/question/1988795

7 0

2 years ago

If you increase the mass of an object and want to move an object a specific distance, what do you need to do

Alexxandr [17]

If you increase the mass of an object and want to move an object a specific distance, then you need to do extra work than the earlier

The total amount of energy transferred when a force is applied to move an object through some distance

Work Done = Force * Displacement

For example, let us suppose a force of 10 N is used to displace an object by a displacement of 5 m then the work done on the object can be calculated by the above-mentioned formula

work done = 10 N ×5 m

=50 N m

Thus, when an object's mass is increased and it is desired to move it a certain distance, more work must be done than previously.

Learn more about work done from here

brainly.com/question/13662169

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3 0

1 year ago

A battery-operated car utilizes a 120.0 V battery with negligible internal resistance. Find the charge, in coulombs, the batteri

fgiga [73]

Answer:

4.29×10⁵ C

Explanation:

From the question,

The energy stored in the battery = Kinetic energy of the car+ Energy needed to make the car climbed the hill+Energy required to exert a force.

E = 1/2mv²+mgh+Fd.................... Equation 1

Where E = Energy stored in the battery, m = mass of the car, v = velocity of the car, h = height of the hill, F = force exerted on the car, d = distance traveled by the car.

But,

d = vt.................... Equation 2

Where v = velocity, t = time.

Substitute equation 2 into equation 1

E = 1/2mv²+mgh+F(vt)................... Equation 3

Given: m = 770 kg, v = 26 m/s, h = 2.15×10² m = 215 m, F = 5.3×10² N = 530 N, t = 1 hour = 3600 s, g = 9.8 m/s²

Substitute into equation 1

E = 1/2(770)(26²)+(770)(9.8)(215)+(530)(26)(3600)

E = 260260+1622390+49608000

E = 51490650 J

Using,

E = qV................. Equation 4

Where q = charge of the battery, V = Voltage.

make q the subject of the equation

q = E/V............... Equation 5

Given: E = 51490650 J, V = 120 V

Substitute into equation 5

q = 51490650/120

q = 429088.75 C

q = 4.29×10⁵ C

7 0

3 years ago

What is the resultant of a pair of one pound forces at right angles to each other?

OleMash [197]

Answer:

$F_R=\sqrt{2} \ pound.force$

Explanation:

Given that there are two force of 1 pound each at right angles to each other.

The from the vector law of addition:

$F_R=\sqrt{F_1^2+F_2^2}$

where:

$F_R=$ resultant force

$F_1\ \&\ F_2$ be the two of the forces to be added.

$F_R=\sqrt{1^2+1^2}$

$F_R=\sqrt{2} \ pound.force$

8 0

3 years ago