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Mazyrski [523]
3 years ago
7

How much heat is needed to raise the temperature of 8g of water by 20oC?

Physics
1 answer:
Elden [556K]3 years ago
7 0
The following information are given in the question:
Mass, M = 8 g
Temperature, T = 20 degree Celsius
Specific heat of water [this value is a constant] C  = 1 c/gc
Heat, Q = ?
The formula for calculating the amount of heat required is given below:
Q = MCT = 8 * 1 * 20 = 160
Therefore, Q = 160 cal. 
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Consider that a ray of light is travelling from glass to water. The refractive index of water is 1.30 (i e n . ., 1.30 w = ) and
Bess [88]

Answer:

\theta_i=49.88^{\circ}

Explanation:

Total internal reflection can happen when light goes from a medium with higher refractive index (in this case, glass) to a medium with lower refractive index (in this case, water).

Snell's Law tells us that n_isin\theta_i=n_rsin\theta_r, where the <em>i</em> stands for incident (in this case, glass) and the <em>r</em> for refracted (in this case, water). We want to know when \theta_r=90^{\circ}, that is, when n_isin\theta_i=n_r, and this happens when the incident angle is:

\theta_i=arcsin(\frac{n_r}{n_i})

Which for our values means:

\theta_i=arcsin(\frac{1.3}{1.7})=arcsin(0.76470588235)=49.88^{\circ}

6 0
3 years ago
What is the potential difference when the current in a circuit is 5mA and resistance is 30 Ohms
Mashcka [7]

<h2>\bf{ \underline{Given:- }}</h2>

\sf• \: The \:  current \:  in \:  a \:  circuit \:  is  \: 5 \: amps.  \: and  \: resistance \:  is \:  30 \:  Ohms.

\\

<h2>\bf{ \underline{To \:  Find :- }}</h2>

\sf{• \:  The  \: Potential  \: Difference. }

\\

\huge\bf{ \underline{ Solution:- }}

\sf According  \: to  \: the  \: question,

\sf•  \: Current \:  (I) = 5  \: Amps.

\sf• \:  Resistance  \: (R) = 30 \:  Ω

\sf{Potential \:  difference  \: means  \: Voltage \: ( V).}

\sf{We \:  know \: that, }

\bf \red{ \bigstar{ \: V = IR }}

\rightarrow \sf V =5 \times 30

\rightarrow \sf V =150

\\

\sf \purple{Therefore, \:  the \:  potential  \: difference  \: is  \: 150  \: v \: .}

4 0
3 years ago
In deep space, there is very little friction. Once they launch a probe into deep space, where there are no external forces actin
Charra [1.4K]

Answer:

move at constant velocity.

Explanation:

Newton's first law (also known as law of inertia) states that:

"when the net force acting on an object is zero, the object will keep its state of rest or if it is moving, it will continue moving at constant velocity".

In the case of the probe, friction in deep space is negligible, therefore when the engine is shut down, there are no more forces acting on the probe: the net force therefore will be zero, so the probe will move at constant velocity.

5 0
3 years ago
Read 2 more answers
You are standing on the SCALE<br> The scale is on the floor.<br> What are the forces on the SCALE?
ehidna [41]

Answer:

The Normal and Gravitational Force

Explanation:

The normal force pushes up and is between the ground and the scale. The gravitational force is the force exerted on the ground.

3 0
3 years ago
Read 2 more answers
Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
3 years ago
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