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marishachu [46]
2 years ago
7

The car starts from a room with a constant acceleration of 5 ms-2. What path will it pass in 6 seconds and what speed will it re

ach?
Physics
1 answer:
Kobotan [32]2 years ago
4 0

Answer: The speed will be  30 m/s .

Explanation:

Given: Initial velocity of the car: u = 0 m/s

Constant Acceleration: a = 5 m/s²

Time: t= 6 seconds

To find: Final velocity(v)

Formula:  v = u+at

Substitute values in the formula, we get

v=  0+(5)(6) m/s

⇒ v= 30 m/s

i.e. Final velocity = 30 m/s

Hence, the speed will be 30 m/s .

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physics A river flows at a speed vr = 5.37 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shorelin
vova2212 [387]

Answer: Vb is the vector  (-5.37m/s,  8.59 m/s), with a module 10.13m/s

then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9

Explanation:

We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.

We have Vr = (5.37m/s, 0m/s)

Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).

The velocity of the boat Vb will be:

Vb = (-c*sin(32). c*cos(32))

Then we should have that:

5.37 m/s - c*sin(32) = 0

c = (5.37/sin(32))m/s = 10.13 m/s

the velocity in the y-axis is:

10.13m/s*cos(32) = 8.59 m/s

So Vb = (-5.37m/s,  8.59 m/s)

the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.

7 0
3 years ago
Paying off your debts will most likely
g100num [7]
I think the answer is d
4 0
2 years ago
Light in the air is incident at an angle to the surface of (12.0 A) degrees on a piece of glass with an index of refraction of (
Orlov [11]

The question is incomplete. You dis not provide values for A and B. Here is the complete question

Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.

A = 12

B = 18

Answer:

18.5⁰

Explanation:

Angle of incidence i = 12.0 + A

A = 12

= 12.0 + 12

= 14

Refractive index u = 1.10 + B/100

= 1.10 + 18/100

= 1.10 + 0.18

= 1.28

We then find the angle of refraction index u

u = sine i / sin r

u = sine24/sinr

1.28 = sine 24 / sine r

1.28Sine r = sin24

1.28 sine r = 0.4067

Sine r = 0.4067/1.28

r = sine^-1(0.317)

r = 18.481

= 18.5⁰

4 0
2 years ago
The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potentia
MAVERICK [17]

Answer:

Energy Lost for group A's car = 0.687 J

Energy Lost for group B's car = 0.55 J

Explanation:

The exact question is as follows :

Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.

To find - How much energy is lost due to heat for group A's car ?

              How much for Group B's car ?

Solution -

We know that,

GPE = 1 Joule (Potential Energy)

Now,

For Group A -

Energy Lost = GPE - KE

                    = 1 J - 0.313 J

                    = 0.687 J

So,

Energy Lost for group A's car = 0.687 J

Now,

For Group B -

Energy Lost = GPE - KE

                    = 1 J - 0.45 J

                    = 0.55 J

So,

Energy Lost for group B's car = 0.55 J

8 0
3 years ago
Two positive charges of same magnitude are separated by some distance if we bring a unit positive charge from one charge to anot
Studentka2010 [4]

Answer:

Increases.

Explanation:

The electric potential increases when the two positive charges of same magnitude bring close to one charge to another because there is repulsive force between them due to same charge and when the two opposite charges move away from each other, the potential energy decreases. When two opposite charges are brought closer together, electric potential energy decreases while on the other hand, when we move opposite charges apart from each other than the work done against the attractive force that leads to an increase in electric potential energy.

3 0
2 years ago
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