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kap26 [50]
2 years ago
14

If the mass of a material is 76 grams and the volume of the material is 14 cm3, what would the density of the material be?

Physics
1 answer:
azamat2 years ago
6 0

Answer:

5.846g/cm³

Explanation:

Density is defined a mass per unit volume of a substance

Density = mass/volume

From the problem:

mass of material = 76g

volume of material = 14cm³

Density = 76g/14cm³

              = 5.846g/cm³

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A net force of 500 N acts on a 100 kg cart. What is the acceleration of the cart if the mass is doubled?
oksian1 [2.3K]

Answer:

2.5 m/s²

Explanation:

You can solve the following equation: F=ma for acceleration.

You'll be left with this:

a=F/m

And then you substitute the force and the doubled mass

a=500N/200kg

a=2.5 m/s²

8 0
2 years ago
A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general
valentina_108 [34]

Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength \lambda_i

• The wave does not satisfy the boundary conditions y_i(0;t)=0


Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions,y(0,1)=0


and y(L,t)=0

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength \lambda_i  can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, y_i(0;t)=0 and y_i(L;t)=0, that are satisfied only for particular value of \lambda_i  .

Given below are the incorrect options about the wave in the string.

•  A_i must be chosen so that the wave fits exactly o the string.

• Any one of  A_i or \lambda_i  or f_i  can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies f_i and the wavelength \lambda_i  must be such that y_i(0;t) = y_i(L;t)=0


(C)

Expression for the wavelength of the various normal modes for a string is,

\lambda_n=\frac{2L}{n} (1)

When n=1 , this is the longest wavelength mode.

Substitute 1 for n in equation (1).

\lambda_n=\frac{2L}{1}\\\\2L

When n=2 , this is the second longest wavelength mode.

Substitute 2 for n in equation (1).

\lambda_n=\frac{2L}{2}\\\\L

When n=3, this is the third longest wavelength mode.

Substitute 3 for n in equation (1).

\lambda_n=\frac{2L}{3}

Therefore, the three longest wavelengths are 2L,L and \frac{2L}{3}.

(D)

Expression for the frequency of the various normal modes for a string is,

f_n=\frac{v}{\lambda_n}

For the case of frequency of the i^{th} normal mode the above equation becomes.

f_i=\frac{v}{\lambda_i}

Here, f_i is the frequency of the i^{th} normal mode, v is wave speed, and \lambda_i is the wavelength of i^{th} normal mode.

Therefore, the frequency of i^{th} normal mode is  f_i=\frac{v}{\lambda_i}

.

6 0
3 years ago
What is the magnite of the net displacement of the mouse?
agasfer [191]

Complete Question

A field mouse trying to escape a hawk runs east for 5.0m, darts southeast for 3.0m, then drops 1.0m down a hole into its burrow. What is the magnitude of the net displacement of the mouse?

Answer:

The  values is  s =  7.49 \  m

Explanation:

From the question we are told that

   The  distance it travels eastward is  x =  5.0 \ m

   The distance it travel towards the southeast  is l  =  3.0\  m

   The distance it travel towards the south is  z =  1 \  m

 

Let x-axis  be east

      y-axis  south

       z-axis into the ground

The angle made between east and south is  \theta  =  45^o

The displacement toward x-axis is

       x =  5 +  3cos(45)

       x =  7.12

 The  displacement toward the y-axis is  

     y  =  3 *  sin (45)

      y =  2.123

Now the overall displacement of the rat is mathematically evaluated as

        s =  \sqrt{7.12^2 +  2.12^2 +  1^2}

      s =  7.49 \  m

     

   

3 0
3 years ago
A force acting on a body causes a change in the momentum of the from 12 kgms-1 to 16 kgms-1 in 0.2 s. calculate the magnitude of
Nonamiya [84]

Answer: Impulse = 4 kgm/s

Explanation:

From the question, you're given the following parameters:

Momentum P1 = 12 kgm/s

Momentum P2 = 16 kgm/s

Time t = 0.2 s

According to second law of motion,

Force F = change in momentum ÷ time

That is

F = (P2 - P1)/t

Cross multiply

Ft = P2 - P1

Where Ft = impulse

Substitute P1 and P2 into the formula

Impulse = 16 - 12 = 4 kgm/s

The magnitude of the impulse is therefore 4 kgm/s.

6 0
3 years ago
Two bodies fall freely from different heights and reach the ground simultaneously. The time of descent for the first body is 1s
jok3333 [9.3K]
The initial height of the first body is given by:
h_1 =  \frac{1}{2}gt^2
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
h_1 =  \frac{1}{2}(9.81 m/s^2)(1 s)^2=4.9 m

The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
h_2 =  \frac{1}{2}(9.81 m/s^2)(2 s)^2=19.6 m

The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.
6 0
3 years ago
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