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3 years ago

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A 4.5cm tall object is placed 28cm in front of a spherical mirror. It is to produce a virtual image that is upright and 3.5cm ta

ll.
a) What type of mirror is used ?
b) Where is the image located ?
c) What is the focal length ?
d) What is the radius of curvature ?

1 answer:

Ostrovityanka [42]3 years ago

7 0

A- convex B- 21.77cm C- 0.01022cm D- 0.02044cm

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A man supports himself and the uniform horizontal beam pulling the rope with a force T.The weights of men and the beam are 883 N

artcher [175]

Answer:

T=502.5N

Ax=171.8N

Explanation:

The computation of the tension T in the rope and the forces exerted by the pin at A is shown below:

vertical forces sum = Ay + Tsin20 + T - 245 - 883 = 0

Now

horizontal forces sum = Ax - Tcos70

Now Moment about B

-Ay × 4.8 + 245 × 2.4 + 883 × 1.8=0

Ay=453.6N

Now substitute in sum of vertical forces T=502.5N

Ax=171.8N

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3 years ago

To understand how to find the velocities of objects after a collision.

trasher [3.6K]

There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = $\frac{v}{2}$

Part C: v_1 = $\frac{v}{3}$ ; v_2 = $\frac{4v}{3}$

Part D: v_1 = v_2 = $\frac{v}{4}$

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = $\frac{1}{2}$ m. $v^{2}$

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

$\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f} )$

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

$v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i} + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}$

$v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}$

For all the collisions, object 2 is static, i.e. $v_{2i}$ = 0

<u>Part A</u>: Both objects have the same mass (m), $v_{1i}$ = v and collision is elastic:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = 0

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.m}{m+m}.v$

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

<u>Part B</u>: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

$m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}$

$v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m.v}{m+m}$

v_1 = v_2 = $\frac{v}{2}$

<u>Part C:</u> Object 1 is 2m, object 2 is m and elastic collision:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = $\frac{2m - m}{2m + m } . v$

v_1 = $\frac{v}{3}$

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.2m}{2m+m}.v$

v_2 = $\frac{4v}{3}$

<u>Part D</u>: Object 1 is m, object is 3m and collision is inelastic:

v_1 = v_2 = $v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m}{m+3m}.v$

v_1 = v_2 = $\frac{v}{4}$

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3 years ago

When comparing two circuits, you note that circuit one has twice the resistance and double the voltage of circuit two. This mean

pochemuha

This means that there is same current flow in both the circuit, or the circuit one has twice the power of circuit two.

According to ohm's law, the resistance is given as

I=V/R

Since the circuit one has twice the voltage, and resistance

I1=I2

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GPE= weight•height= 15 N• 0.22meter= 3.3 Joules
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Answer:

Add items to the box

Explanation: I did the test

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