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photoshop1234 [79]
3 years ago
11

At the bottom of the ocean, a rock has a mass 25 g. At sea level, the same rock will have a mass of: less than 25 g, more than 2

5 g, or the same as 25 g?
Physics
2 answers:
storchak [24]3 years ago
5 0
It will be same 25 g
ivann1987 [24]3 years ago
5 0

Answer:

Option (2)

Explanation:

The weight of an object is always more at the sea level than at the base of the ocean floor. The water always forces an upward thrust to an object, as a result of which the weight of the object gets reduced by a certain amount. This property of water generating an upward force is known as the the buoyancy.

Thus, if an object weighs 25 grams at the ocean floor, then at the mean sea level, the weight is slightly more than 25 grams.

Hence, the correct answer is option (2).

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According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave lengt
scZoUnD [109]

Answer:1.55 times

Explanation:

Given

First wavelength(\lambda _1)=450 nm

Second wavelength(\lambda _2)=700 nm

According wien's diplacement law

\lambda T=constant

where \lambda =wavelength

T=Temperature

Let T_1 and T_2 be the temperatures corresponding to \lambda _1 & \lambda _2 respectively.

\lambda _1\times T_1=\lambda _2\times T_2

\frac{T_1}{T_2}=\frac{\lambda _2}{\lambda _1}

\frac{T_1}{T_2}=\frac{700}{450}=1.55

Thus object with \lambda 450 nm is 1.55 times hotter than object with wavelength \lambda =700 nm

8 0
3 years ago
The equation of a transverse wave traveling along a very long string is y 6.0 sin(0.020px 4.0pt), where x and y are expressed in
zhannawk [14.2K]

Answer:

given

y=6.0sin(0.020px + 4.0pt)

the general wave equation moving in the positive directionis

y(x,t) = ymsin(kx -?t)

a) the amplitude is

ym = 6.0cm

b)

we have the angular wave number as

k = 2p /?

or

? = 2p / 0.020p

=1.0*102cm

c)

the frequency is

f = ?/2p

= 4p/2p

= 2.0 Hz

d)

the wave speed is

v = f?

= (100cm)(2.0Hz)

= 2.0*102cm/s

e)

since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction

f)

the maximum transverse speed is

umax =2pfym

= 2p(2.0Hz)(6.0cm)

= 75cm/s

g)

we have

y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]

= -2.0cm

6 0
3 years ago
On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the
IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = \frac{1}{2}\varepsilon _{0}E^{2}

          =\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119

          = 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = \frac{B^{2}}{2\mu _{0}}

=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

8 0
3 years ago
A rocket on Earth experiences an upward applied force from its thrusters. As a result of this force, the rocket accelerates upwa
gayaneshka [121]

Answer:

F=m(11.8m/s²)

For example, if m=10,000kg, F=118,000N.

Explanation:

There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

F-mg=ma

Solving for the force F, we obtain that:

F=ma+mg=m(a+g)

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

F= m(2m/s^{2}+9.8m/s^{2})\\\\F=m(11.8m/s^{2})

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

F=(10,000kg)(11.8m/s^{2})\\\\F=118,000N

In this case, the force from the rocket's thrusters is equal to 118,000N.

5 0
3 years ago
1. A man throws a ball up with a velocity of 30 m/s. How high does it get?
ElenaW [278]
Yes the answer is g=5
8 0
3 years ago
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