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mafiozo [28]
2 years ago
9

Question 81 point)

Physics
1 answer:
azamat2 years ago
4 0

Answer:

 (C) Mechanical Force

Explanation:

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Which statement is true if the refractive index of medium A is greater than that of medium B?
allsm [11]

Answer:

I think it is C

Explanation:

7 0
3 years ago
You walk towards Starbucks southward for 14 meters and then west for 9 meters. The trip takes you 3 minutes to walk there. What
zimovet [89]

Although you walked (14+9)= 23 meters of distance, you ended up only 16.64 meters from where you started. That's your "displacement".

Velocity = (displacement / (time)

Velocity = (16.64m)/(180s)

Velocity = 0.0925 m/s roughly Southwest.

(That's about 3.6 INCHES per second. Apparently, you really NEED that mocha java and it's associated drug.)

8 0
3 years ago
An engineer can increase the magnitude of the magnification of a compound microscope by
jok3333 [9.3K]

Answer:

Option B

Explanation:

Magnification of Microscope is  

M = M_o \times M_e

Mo= Magnification of objective lens and

Me= magnification of the eyepiece.  

Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.  

Magnification,  

M\ \alpha\ \dfrac{1}{f}

when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.

Thus. Magnification will increase by decreasing the focal length.

The correct answer is Option B i.e. using shorter focal length

6 0
3 years ago
8)
marishachu [46]

Answer:

The box will experience an acceleration.

Explanation:

Here, 2 N and 3 N forces are acting opposite to each other. In this case, the net force experience by the box would be (3-2)N = 1 N towards right. Since acceleration is directly proportional to the net force, therefore the box will experience an acceleration.

6 0
3 years ago
A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface
Fiesta28 [93]

Answer:

X_2=25.27m

Explanation:

Here we will call:

1. E_1: The energy when the first spring is compress

2. E_2: The energy after the mass is liberated by the spring

3. E_3: The energy before the second string catch the mass

4. E_4: The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. E_1 = E_2

2. E_2 -E_3= W_f

3.E_3 = E_4

where W_f is the work of the friction.

1. equation 1 is equal to:

\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2

where K is the constant of the spring, x is the distance compressed, M is the mass and V_2 the velocity, so:

\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2

Solving for velocity, we get:

V_2 = 65.319 m/s

2. Now, equation 2 is equal to:

\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd

where M is the mass, V_2 the velocity in the situation 2, V_3 is the velocity in the situation 3, U_k is the coefficient of the friction, N the normal force and d the distance, so:

\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2)

Volving for V_3, we get:

V_3 = 65.27 m/s

3. Finally, equation 3 is equal to:

\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2

where K_2 is the constant of the second spring and X_2 is the compress of the second spring, so:

\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2

solving for X_2, we get:

X_2=25.27m

3 0
3 years ago
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