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timofeeve [1]
2 years ago
12

If 90J of energy are available for every 30C of charge, what is the potential difference?

Physics
1 answer:
Natasha2012 [34]2 years ago
5 0

Answer:

3 v

Explanation:

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A car traveling at 5m/s starts to speed up after 3 seconds its velocity has increased to 11 m/s what is its acceleration
vfiekz [6]

Answer:

a=(v-u)/t

Explanation:

a =(11-5)/3

a= 8/3

a= 2.6 m/s

4 0
2 years ago
Please help
Hitman42 [59]

Answer:

3200 m/s

Explanation:

The speed of Sound through Argon is 319.

The speed of sound in any chemical element in the fluid phase has one temperature-dependent value. In the solid phase, different types of sound wave may be propagated, each with its own speed: among these types of wave are longitudinal (as in fluids), transversal, and (along a surface or plate) extensional.

If argon could exist as solid, then 3200 m/s is the best speed.

7 0
3 years ago
Read 2 more answers
A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round
Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s

v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s

Now we find the centripetal acceleration which is given by

a_c=\frac{v^2}{r}

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2

we also have a tangential acceleration, which is given by

a_t = \frac{v-u}{t}

where

t = 17.0 s

Substituting values,

a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2

6 0
3 years ago
Read 2 more answers
A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio
Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

\sum F_{x}=0

we can see there are only two forces in x, which are the weight on x and the static friction, so:

-W_{x}+f_{s}=0

when solving for the static friction we get:

f_{s}=W_{x}

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

W_{x}=mg sin \alpha

we can substitute this on our sum of forces equation:

f_{s}=mg sin \alpha

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

f_{s}=N\mu_{s}

so we can substitute this on our equation:

N\mu_{s}=mg sin \alpha

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

\sum F_{y}=0

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

N-W_{y}=0

when solving for N we get:

N=W_{y}

When seeing the free-body diagram we can determine that:

W_{y}=mg cos \alpha

so we can substitute that in the sum of y-forces equation, so we get:

N=mg cos \alpha

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

mg cos \alpha \mu_{s}=mg sin \alpha

we can divide both sides of the equation into mg so we get:

cos \alpha \mu_{s}=sin \alpha

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for \mu_{s} we get:

\mu_{s}=\frac{sin \alpha}{cos \alpha}

when simplifying this we get that:

\mu_{s}=tan \alpha

now we can solve for the angle so we get:

\alpha= tan^{-1}(\mu_{s})

and we can substitute the given value so we get:

\alpha= tan^{-1}(0.350)

which yields:

α=19.29°

which rounds to:

α=19°

8 0
3 years ago
A 0.59 kg bullfrog is sitting at rest on a level log. how large is the normal force of the log on the bullfrog?
ladessa [460]
<span>The bullfrog is sitting at rest on the log. The force of gravity pulls down on the bullfrog. We can find the weight of the bullfrog due to the force of gravity. weight = mg = (0.59 kg) x (9.80 m/s^2) weight = 5.782 N The bullfrog is pressing down on the log with a force of 5.782 newtons. Newton's third law tells us that the log must be pushing up on the bullfrog with a force of the same magnitude. Therefore, the normal force of the log on the bullfrog is 5.782 N</span>
7 0
2 years ago
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