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zmey [24]
3 years ago
9

What force pulls the truck/ramp down towards the track? Group of answer choices

Physics
1 answer:
Effectus [21]3 years ago
8 0

Answer:

B.Gravity

Explanation:

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Without an atmosphere, the equatorial curve would show minimum daily values on the solstices in June when the sub-solar point is
lapo4ka [179]

Without an atmosphere, the equatorial curve would show minimum daily values on the solstices in June when the sub-solar point is located at 23.5°N and in December when the sub-solar point is at 23.5°S latitude.

Explanation:

At the sub-solar point, the sun strikes directly at the surface with an angle of 90 degrees at a given point.

Solistice refers to that point in time when the sun’s zenith is located at the farthest point from the equator.  

During summer solistice on June 21, the sun’s zenith reaches northernmost point, sub-solar point is fixed at 23.5°S Tropic of Cancer making the earth tilt 23.4 degrees

During winter soliscitse on December 21, the sub-solar point is fixed at)  Tropic of Capricorn.

3 0
3 years ago
In 40 words or fewer, summarize the connection between kinetic and thermal energy.<br> PLEASE REPLY
xz_007 [3.2K]

Answer:

Explanation:

kenetic is made for thermal things

8 0
3 years ago
A man pushes on a trunk with a force of 250 newtons. The trunk does not move. How much positive work is done on the trunk?
Digiron [165]

Answer:

F is 250 N

d is 0 m

F x d

=250 x 0

=0

The answer is 0.0 J.

3 0
3 years ago
A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end
Natalka [10]

Answer:

The angular acceleration α = 14.7 rad/s²

Explanation:

The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod  and L = length of rod = 4.00 m. α = angular acceleration of rod

Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.

So Iα = Wr

Substituting the value of the variables, we have

mL²α/12 = mgL/2

Simplifying by dividing through by mL, we have

mL²α/12mL = mgL/2mL

Lα/12 = g/2

multiplying both sides by 12, we have

Lα/12 × 12 = g/2 × 12

αL = 6g

α = 6g/L

α = 6 × 9.8 m/s² ÷ 4.00 m

α = 58.8 m/s² ÷ 4.00 m

α = 14.7 rad/s²

So, the angular acceleration α = 14.7 rad/s²

5 0
3 years ago
An electric toy with a resistance of 2.50 Ω is operated by a 3.00-V battery. (a) What current does the toy draw? (b) Assuming th
Leno4ka [110]

Answer:

a) The current is i = 1.2 A

b) The charge is Q = 17280 C

c) The energy is E = 43200 J

Explanation:

a) The current is given by the ohm's law wich is:

i = V/R = 3/2.5 = 1.2 A

b) Since the charge is steady we can use the following equation to find the charge amount in that time:

i = Q/t

Q = t*i

Where t is in seconds, so we have 4h * 3600 = 14400 s

Q = 1.2*14400 = 17280 C

c)  The energy is the power delivered to the toy multiplied by the time:

P = 1.2*2.5 = 3 W

E = P*t = 3*14400 = 43200 J

7 0
3 years ago
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