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3 years ago

Will plane flat mirror will make a real or virtual image? Real Virtual

1 answer:

4 0

Image formed by a plane mirror is always virtual which means that the light rays do not actually come from the image but upright and these of the same shape and size are the object it is reflecting.

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In which direction does a bag at rest move when a force of 20 newtons is applied from the right? A. in the direction of the appl

-Dominant- [34]

The correct answer is A. In the direction of applied force. This is because acceleration occurs n the direction of applied force according to Newtons second law of motion which states that the acceleration of a body is directly proportional to the applied force and takes place in the direction of force.

5 0

3 years ago

Read 2 more answers

Please help me, this is a physics test.

sweet-ann [11.9K]

Answer:

a = 2 [m/s²]

Explanation:

To be able to solve this problem we must make it clear that the starting point when the time is equal to zero, the velocity is 5 [m/s] and when three seconds have passed the velocity is 11 [m/s], this point is the final point or the final velocity.

We can use the following equation.

$v_{f}=v_{o}+a*t\\$

where:

Vf = final velocity = 11 [m/s]

Vo = initial velocity = 5 [m/s]

a = acceleration [m/s²]

t = time = 3 [s]

$11 = 5 + a*3\\6=3*a\\a= 2[m/s^{2} ]$

4 0

3 years ago

A 65.8-kg person throws a 0.0413 kg snowball forward with a ground speed of 32.5 m/s. A second person, with a mass of 58.7 kg, c

guapka [62]

Answer:

$v_{1} = 2.490\,\frac{m}{s}$

$v_{2} = 0.023\,\frac{m}{s}$

Explanation:

The statement is described physically by means of the Principle of Momentum Conservation. Let assume that first person moves in the positive direction:

First Person

$(65.8\,kg)\cdot (2.51\,\frac{m}{s}) = (65.8\,kg)\cdot v_{1} + (0.0413\,kg)\cdot (32.5\,\frac{m}{s} )$

Second Person

$(0.0413\,kg)\cdot (32.5\,\frac{m}{s})+(58.7\,kg)\cdot (0\,\frac{m}{s})=(0.0413\,kg+58.7\,kg)\cdot v_{2}$

The final velocities of the two people after the snowball is exchanged is:

$v_{1} = 2.490\,\frac{m}{s}$

$v_{2} = 0.023\,\frac{m}{s}$

6 0

3 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0

3 years ago

Given a wire with a cross section of .45cm^2, a length of 3cm, and an internal resistance of 3 ohms, show your 5 steps to solve

Gre4nikov [31]

Resistance ∞ (proportional) length
resistance ∞ 1/ area

therefore,
(the constant that we take is known as the resistivity)

resistance = (resistivity*length )/ area
resistivity = (resistance * area ) / length
= (3 * 45) / 3 = 135/3 = 45 Ωm

in short your answer is 45 Ωm

4 0

3 years ago