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OLga [1]
3 years ago
7

On another planet, a morble is released from rest at the top of a high cliff. It falls 4.00 m in the first 1 s of its motion. Th

rough what additional distance does it fall in the next 1 s?
(a) 4.00 m
(b) 8.00 m
(c) 12.0 m
(d) 16.0 m
(e) 20.0 m
Physics
1 answer:
BabaBlast [244]3 years ago
6 0

Answer:

option  c is correct

additional distance is 12 m

Explanation:

given data

height = 4 m

time = 1 s

to find out

additional distance in next 1 s

solution

we will apply here formula for height

height = 1/2 × g×t²       ...................1

put here value height = 4 and t = 1 s

4 = 1/2 × g×(1)²

g = 8 m/s²

and

for time next 1 sec

from equation 1

height = 1/2 × g×t²  

put here t = 2 s so

h2 - h1 = 1/2 × g×(2²  - 1² )

h2 - h1 = 1/2 × 8×(2²  - 1² )

h2 - h1 = 4 ( 4-1)

h2 - h1 = 12 m

so option  c is correct

additional distance is 12 m

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A cylindrical solenoid (radius r=0.6 m, turns N=600, length l=0.5 m) carries 15 A of current. What is the inductance L of the co
Anuta_ua [19.1K]

Answer:

L = 1.023 H

Explanation:

given,

radius of the cylindrical solenoid = r = 0.6 m

Number of turns = N = 600

Length = l = 0.5 m

Current in the cylindrical solenoid = 15 A

Inductance in the coil = ?

using formula

L = \dfrac{\mu_0\ N^2\ A}{l}

L = \dfrac{4 \pi \times 10^{-7}\times 600^2\times \pi r^2}{0.5}

L = \dfrac{4 \pi \times 10^{-7}\times 600^2\times \pi 0.6^2}{0.5}

L = \dfrac{4 \pi \times 10^{-7}\times 600^2\times \pi 0.6^2}{0.5}

L = 1.023 H

the inductance L of the coil is = 1.023 H

8 0
3 years ago
If this standing wave is 48.8 meters long from end to end, what is the wavelength?
Mrrafil [7]

The wavelength of a standing wave is 8.13 m.

<h3>What is a wavelength?</h3>

The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.

The wavelength is also defined as the distance between two locations in a wave that have the same oscillation phase.

The given data in the problem is;

String length(L)= 48.8 metere

Wavelength(λ)=?

The length of the wave having n nodes is found as;

L=nλ

Substitute the given value;

48.8 = 6λ

λ= 8.13 m

Hence, the wavelength of a standing wave is 8.13 m.

To learn more about the wavelength, refer to the link;

brainly.com/question/7143261

#SPJ1

6 0
2 years ago
Suppose it takes a constant force a time of 6.0 seconds to slow a 2500 kg truck
erastovalidia [21]

Answer:

3.3\cdot 10^3\:\mathrm{N}

Explanation:

Impulse on an object is given by \mathrm{[impulse]}=F\Delta t.

However, it's also given as change in momentum (impulse-momentum theorem).

Therefore, we can set the change in momentum equal to the former formula for impulse:

\Delta p=F\Delta t.

Momentum is given by p=mv. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:

p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}.

Now we plug in our values and solve:

\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}(two significant figures).

6 0
3 years ago
SUPPOSE THAT YOURE FACING A STRAIGHT CURRENT CARRYING CONDUCTOR, AND the current isfowing towards you. The lines of magnetic for
enot [183]
According to the right hand grip rule; if you point your right thumb in the direction of the current flow and curl your fingers around the current carrying conductor, the fingers will point in the direction of the circular magnetic field around the conductor. Therefore,if the the current carrying carrying conductor is held with the current flowing towards you, using right hand grip rule the lines of magnetic field will act towards a counterclockwise direction or anticlockwise direction.
7 0
3 years ago
A solid conducting sphere with radius RR that carries positive charge QQ is concentric with a very thin insulating shell of radi
svetlana [45]

Answer:

E=0 at r < R;

E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}} at 2R > r > R;

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}} at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon} (integral over a closed surface)

where,

E = Electric field

Q_{enclosed} = charged enclosed within the closed surface

\epsilon = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

Q_{enclosed} = 0 and hence E = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

Q_{enclosed} = Q,

therefore,

E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}      

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to 4\pi r^{2})

or, E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}

at r >= 2R

Q_{enclosed} = 2Q

Hence, by similar calculations, we get,

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}

4 0
4 years ago
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