Kinetic energy problemset

Nancy ran a distance of 5km in 30 minutes. What is her speed in meters(m) per hour)h)

Virty [35]

The speed of Nancy who ran a distance of 5 km in 30 minutes is 10000 meters(m) per hour(h).

<h3>What is the rate of speed?</h3>

The rate of speed is the rate at which the total distance is travelled in the time taken. Rate of speed can be given as,

$s=\dfrac{d}{t}$

Here, (d) is the distance travelled by the object and (t) is the time taken but the object to cover that distance.

Nancy ran a distance of 5 km in 30 minutes. There are 60 minutes in one hour. Thus, the time in hour for which Nancy ran is,

$t=\dfrac{30}{60}\\t=0.5\rm\; hr$

The meters in 5 kilometers is,

$d=5\times1000\\d=5000\rm\; m$

She ran 5000 kilometers in 0.5 hours. Thus, the speed of her is,

$s=\dfrac{5000}{0.5}\\s=10000\rm\; km/hr$

Thus, the speed of Nancy who ran a distance of 5 km in 30 minutes is 10000 meters(m) per hour(h).

Learn more about the rate of speed here:

brainly.com/question/359790

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7 0

2 years ago

06

Novosadov [1.4K]

D

Explanation:

The are the same but difference

8 0

2 years ago

A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

B)

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

8 0

3 years ago

Elaborate on the reason(s) that matter is said to move even as in a solid state.

Mariulka [41]

<u>Answer:</u>

The matter does not move in solid state but vibrates.

<u>Explanation:</u>

The atoms inside the matter cannot move or shift their positions without any external force but makes some small vibration movements. Generally in solids, the particles are bound by the attractive forces acting in between the atoms inside the matter.

The small vibrations that are happening inside the matter are because of the external factors like temperature. The increase in temperature raises the kinetic energy of the atoms inside and makes them move faster and this results in the vibration of the matter.

6 0

3 years ago

Read 2 more answers

A 2-kg cart, traveling on a horizontal air track with a speed of 3m/s, collides with a stationary 4-kg cart. The carts stick tog

ladessa [460]

Answer:

The impulse exerted by one cart on the other has a magnitude of 4 N.s.

Explanation:

Given;

mass of the first cart, m₁ = 2 kg

initial speed of the first car, u₁ = 3 m/s

mass of the second cart, m₂ = 4 kg

initial speed of the second cart, u₂ = 0

Let the final speed of both carts = v, since they stick together after collision.

Apply the principle of conservation of momentum to determine v

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 3 + 0 = v(2 + 4)

6 = 6v

v = 1 m/s

Impulse is given by;

I = ft = mΔv = m(

The impulse exerted by the first cart on the second cart is given;

I = 2 (3 -1 )

I = 4 N.s

The impulse exerted by the second cart on the first cart is given;

I = 4(0-1)

I = - 4 N.s (equal in magnitude but opposite in direction to the impulse exerted by the first).

Therefore, the impulse exerted by one cart on the other has a magnitude of 4 N.s.

8 0

3 years ago