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Cerrena [4.2K]
3 years ago
10

A block of mass 20 kg sits on a ramp with an angle of 31 degrees above the horizontal. Assuming no friction, how fast will the b

lock accelerate down the ramp?
7.89 m/s^2
5.34 m/s^2
2.98 m/s^2
4.17 m/s^2
Physics
2 answers:
Artyom0805 [142]3 years ago
4 0

Answer:

2.98 m/s^2

Explanation:

I have done this before and it was a question on my physics test

snow_lady [41]3 years ago
4 0

Answer:

3ird one my boy ive seen alot of problems like this

Explanation:

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16)
VMariaS [17]

Answer:

a.proton, proton

hope this helped :) have a goodday

6 0
3 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
iren [92.7K]

Answer:

372.3 J/^{\circ}C

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

P=VI=(3.6)(2.6)=9.36 W

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

E=Pt=(9.36)(350)=3276 J

Finally, the change in temperature of an object is related to the energy supplied by

E=C\Delta T

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C is the change in temperature

Solving for C, we find:

C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

5 0
2 years ago
I don’t know this one I need help
nikdorinn [45]
D
Using the kinetic energy 1/2mv^2 formula
5*10^5 is the answer
4 0
2 years ago
A fully loaded cart with a mass of 2200 kg starts from the top of a 12-meter hill on a roller coaster.
Salsk061 [2.6K]

Answer:

A. potential energy is 258720 Joule

Explanation:

A.Gravitational potential energy is: PE = m × g × h

velocity =  15.33 m/s when the car reaches the bottom of the hill.

where, m = mass

            g = acceleration due to gravity

            h = height from the bottom of hill.

The potential energy is : m×g×h

                                     =(2200×9.8×12)

                                     =258720 Joule

B. at the bottom of the hill, the potential energy is converted into kinetic energy so PE at top = KE at bottom

                    kinetic energy= \frac{1}{2}(m*v^{2})

where v = velocity

          m= mass

therefore,               v=\sqrt\frac{2*K.E}{m} {}

                         or,  v=\sqrt{\frac{2*258720}{2200} }

                         or,   v=15.33 m/s

7 0
2 years ago
A hoodlum throws a stone vertically downward with an initial speed v0 from the roof of a building, a height h above the ground.
Vaselesa [24]
V2=u2+2as
v2=144+600
v2=744
v=√744
6 0
3 years ago
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