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kykrilka [37]
3 years ago
8

A certain car battery with a 12.0 V emf has an initial charge of 131 A · h. Assuming that the potential across the terminals sta

ys constant until the battery is completely discharged, for how many hours can it deliver energy at the rate of 130 W?
Physics
1 answer:
irga5000 [103]3 years ago
7 0

Answer:

The battery can supply 130 W for 11.75 h

Explanation:

In order to discover the time in wich the battery can supply this energy we need to find how much current is being drawn from it, we do that by using the equation for real power that is P = V*I, since we have V and P we can solve for I as seen bellow:

I = P/V = 130/12 = 10.834 A

We can use this value to find how many hours the power can supply said current. We do that by dividing the current capacity of the battery by the current drawn:

t = 141/12 = 11.75 h

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98 POINTS FOR ANYONE THAT DOES THIS NOW!!
Varvara68 [4.7K]

In simpler terms, a proton and neutron weigh 1 amu (atomic mass unit) each.

The nucleus has 15 protons and 18 neutrons. Since a proton's and a neutron's weight is only 1 amu, we can simply add the number of protons and neutrons to find the total mass of the nucleus:

15 + 18 = \boxed{33}

The nucleus' mass is 33 amu.

8 0
3 years ago
Read 2 more answers
A car moving in a straight line starts at x = 0 at t = 0 . It passes the point x = 30.0 m with a speed of 10.0 m/s at t = 3.00 s
Bogdan [553]

Answer:

a) v=20.3m/s

b) a=2.35m/s^2

Explanation:

From the exercise we know:

x_{1}=30m; v_{1}=10m/s; t_{1}=3s

x_{2}=375m; v_{2}=50m/s; t_{2}=20s

The formula for average velocity is:

v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}

a) v=\frac{375m-30m}{20s-3s}=20.3m/s

The formula for average acceleration is:

a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}

b) a=\frac{(50-10)m/s}{(20-3)s}=2.35m/s^2

4 0
3 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
3 years ago
A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from
Alenkasestr [34]

Answer:

V₁  = 5.6 m/s

V₂ = 7.2 m/s

V₃ = 8.8 m/s

Explanation:

Average velocity: Average velocity can be defined as the ratio of the total  displacement to the total time taken. The S.I unit of Average velocity is m/s.

For the first 2 s,

V₁ = Δd₁/t

Where V₁  = Average velocity for the first 2 s

Where Δd₁= distance, t = time

Δd₁ = 25.6-14.4 = 11.2 m t = 2 s

V₁ = 11.2/2

V₁ = 5.6 m/s

For the second 2 s,

V₂ =Δd₂/t

Where V₂ = average velocity for the second 2 s.

Δd₂= 40-25.6 = 14.4 m, t= 2 s

V₂ = 14.4/2

V₂ = 7.2 m/s

For the last 2 seconds,

V₃ =Δd₃/t

Where V₃ = average velocity for the last 2 s

where Δd₃ = 57.6- 40 = 17.6 m, t = 2 s

V₃ = 17.6/2

V₃ = 8.8 m/s.

8 0
4 years ago
A point charge q1 = 4.50 nC is located on the x-axis at x = 1.95 m , and a second point charge q2 = -6.80 nC is on the y-axis at
Vinvika [58]

Answer:

Explanation:

One charge is situated at x = 1.95 m . Second charge is situated at y = 1.00 m

These two charges are situated outside sphere as it has radius of .365 m with center at origin. So charge inside sphere = zero.

Applying Gauss's theorem

Flux through spherical surface = charge inside sphere / ε₀

= 0 / ε₀

= 0 Ans .

3 0
3 years ago
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