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Soloha48 [4]
3 years ago
10

If the mass of a moving object decreases from 100 grams to 25 grams, what happens to its momentum

Physics
1 answer:
cluponka [151]3 years ago
5 0
The momentum will become 1/4 as much.
0.100 v === 0.25 v 
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Find the wavelength λ of the 80.0-khz wave emitted by the bat. express your answer in millimeters.
vfiekz [6]

Answer:

4.29 millimeters

Explanation:

Bats emit ultrasound waves: in air, ultrasound waves travel at a speed of

v=343 m/s

The frequency of the waves emitted by this bat is:

f=80.0 kHz = 80,000 Hz

Therefore we can find the wavelength of the wave emitted by the bat by using the relationship between speed, frequency and wavelength:

\lambda=\frac{v}{f}=\frac{343 m/s}{80,000 Hz}=4.29\cdot 10^{-3} m=4.29 mm

4 0
3 years ago
Two masses are attracted by a gravitational force of 7 N.
damaskus [11]

Answer:

The answer to your question is: F  = 0.4375 N. The force will be 16 times lower than with the first conditions.

Explanation:

Data

F = 7 N

F = ?  if the masses is quartered

Formula

F = \frac{Km1m2}{r2}

Process

Normal conditions F = Km₁m₂/r²  = 7              

When masses quartered        F = K(m₁/4)(m₂/4)/r²  = ?

                                                F = K(m₁m₂/16)/r²

                                                F = K(m₁m₂/16r²      = 7/16  = 0.4375 N

3 0
3 years ago
What do you think we call this graphical representation based on your prior experience with electric fields and electric field l
slavikrds [6]

Answer:

Explanation:

The strengthcompassion field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area perpendicular to the lines. The direction of the electric field is tangent to the field line at any point in space. Field lines can never cross. These pattern of lines, sometimes referred to as electric field lines, point in the direction that a positive test charge would accelerate if placed upon the line. As such, the lines are directed away from positively charged source charges and toward negatively charged source charges.

Rules for drawing electric field lines

1. Electric field lines are always drawn from High potential to

low potential.

2. Two electric field lines can never intersect each other.

3. The net electric field inside a Conductor is Zero.

4. Electric field line from a positive charge is drawn radially outwards and from a negative charge radially inwards.

5. The density of electric field lines tells the strength of the electric field at that region.

6. Electric field lines terminate Perpendicularly to the surface of a conductor.

A vector quantity has a direction and a magnitude, while a scalar has only a magnitude. You can tell if a quantity is a vector by whether or not it has a direction associated with it.

So, electric fields are vector quantity due to the fact any student can tell you that a compass is used to determine which direction is north.

Since the compass always point northward, then it has a direction and magnitude and so it is a vector quantity

6 0
3 years ago
A 200g block on a 50cm long string swings in a circle, it's frictionless and 75rpm. What is its speed and tension on string
krek1111 [17]
Angular velocity = (75x2pie)/60
                          =2.5pie ras^-1 
linear velocity(or speed) at end of string, v = radius x angular velocity
                                                           v= 0.5 x 2.5pie
                                                           v=3.93 ms^-1

tension of string (I beleve is centeral force aplied by string), F= (mv^2)/r
                                                                                      F= (0.2 x 3.93^2)/0.5
                                                                                      F=6.18 N
(sorry if wrong)
3 0
3 years ago
Read 2 more answers
An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 14.0
Reptile [31]

Answer:

28 m/s^2

Explanation:

distance, s = 14 m

time, t = 2 - 1 = 1 s

initial velocity, u = 0 m/s

Let a be the acceleration.

Use third equation of motion

s = ut + \frac{1}{2}at^{2}

14 = 0 + \frac{1}{2}a\times 1^{2}

a = 28 m/s^2

Thus, the acceleration is 28 m/s^2.

7 0
3 years ago
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