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Arte-miy333 [17]
3 years ago
13

A crew member is walking on a tugboat that is pulling a barge. The tugboat is moving at a constant speed upstream in a river tha

t has a constant downstream current. The velocity of the crew member with respect to the tugboat is v₁. What is the velocity of the crew member with respect to the barge?
Physics
1 answer:
Marianna [84]3 years ago
4 0

Answer:

v₁

Explanation:

A crew member is walking on a tugboat that is pulling a barge. The tugboat is moving at a constant speed upstream in a river that has a constant downstream current. The velocity of the crew member with respect to the tugboat is v₁. What is the velocity of the crew member with respect to the barge?

velocity is the change in displacement per time.

the velocity of the crew member with respect to the barge is v₁ since the barge is moving in the velocity of tugboat

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A light with a second-order bright band forms a diffraction angle of 30. 0°. The diffraction grating has 250. 0 lines per mm. Wh
Luden [163]

The distance between two successive troughs or crests is known as the wavelength. The wavelength of the light will be 1000 nm.

How do you define wavelength?

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.

Diffraction angle= 30⁰

Diffraction grating per mm= 250

wavelength = ?

Mathematically the equation of bright band is given by

\rm \lambda= \frac{sin\theta}{nN}

\rm \lambda= \frac{sin23^0}{250\times 2}

\rm \lambda= 0.000001 m

\rm \lambda= 1000 nm

Hence the wavelength of the light will be 1000 nm.

To learn more about the wavelength refer to the link;

brainly.com/question/7143261

8 0
2 years ago
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A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A
Sphinxa [80]

Answer:

Density of 127 I = \rm 1.79\times 10^{14}\ g/cm^3.

Also, \rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

Explanation:

Given, the radius of a nucleus is given as

\rm r=kA^{1/3}.

where,

  • \rm k = 1.3\times 10^{-13} cm.
  • A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.

For the nucleus 127 I,

Mass, M = \rm 2.1\times 10^{-22}\ g.

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.

On comparing with the density of the solid iodine,

\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

7 0
3 years ago
What is the study of energy and how it affects matter called?
bogdanovich [222]
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We push a 38.4-kg box across the floor at constant velocity. If we are pushing a horizontal force of 238 N, find the coefficient
zmey [24]

Answer: 0.62

Explanation:

Coefficient of friction is defined as the ratio of the moving force (Fm) acting on a body to the normal reaction (R).

Note that the normal reaction acts vertically on the object and is equal to the objects weight (W) i.e W=R

Since W = mg, W = 38.4 ×10

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Normal reaction = 384N

The horizontal force acting on the body will be the moving force which is 238N

Coefficient of friction = Fm/R

Coefficient of friction = 238/384

Coefficient of friction = 0.62

Therefore, coefficient of kinetic friction between the box and the floor is 0.62

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A pulled tablecloth exerts a frictional force of 0.6 n on a plate with a mass of 0.4 kg. what is the acceleration of the plate?
Ludmilka [50]
The formula for acceleration is a = F/m; Where: F = force; m = mass
Given: F = .6n; m = .4kg; a = ?
a = F/m
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Therefore, the acceleration of the plate is 1.5 m/s^2
6 0
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