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NemiM [27]
3 years ago
10

Define specific heat capacity

Physics
2 answers:
sleet_krkn [62]3 years ago
7 0

Answer:you can just look this up yknow?

Explanation:

kirza4 [7]3 years ago
4 0
The heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).
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You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If
Stells [14]

Answer:

Komila should sit 0.33m from the middle of the board towards tahreen.

Explanation:

We are told to treat each student as point-like objects. So i have attached a rigid body diagram to depict this.

From the diagram,

F_d is force exerted by dan

F_t is force exerted by tahreen

F_k is force exerted by komila

F_b is force of board at the mid point.

x1 is distance of dan from the centre of the chair

x2 is distance of komila from the centre of the chair

x3 is distance of tahreen from komila

We are given;

Mass of Dan;m_d = 62 kg

Mass of tahreen;m_t = 50 kg

Mass of komila;m_k = 54 kg

Now, taking moments about the centre of the chair, we have;

(F_d*x1) - (F_k*x2) - (F_t(x2 + x3)) = 0

Now,F_d = m_d*g ; F_t = m_t*g ; F_k = m_k*g

We are told that the board is 3m long. So, if we assume that the fulcrum position of the chair coincides with the midpoint of boards length, we'll have;

x1 = (x2 + x3) = 1.5

Thus, we now have;

(F_d*1.5) - (F_k*x2) - (F_t*1.5) = 0

F_d = m_d*g = 62 * 9.8 = 607.6 N

F_t = m_t * g = 50 x 9.8 = 490 N

F_k = m_k * g = 54 x 9.8 = 529.2 N

So plugging in these values, we have;

(607.6 * 1.5) - (529.2 * x2) - (490 * 1.5) = 0

911.4 - 735 = 529.2 x2

529.2 x2 = 176.4

x2 = 176.4/529.2

x2 = 0.33m

Komila should sit 0.24m from the middle of the board towards tahreen

7 0
3 years ago
What must be the distance between point charge q1 = 28.0 μC and point charge q2 = −57.0 μC for the electrostatic force between t
vlabodo [156]

Answer:

1.686 m

Explanation:

From coulomb's law,

F = kq1q2/r² ...................................... Equation 1

Where F = electrostatic force  between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.

making r the subject of the equation,

r = √(kq1q2/F).......................... Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

Constant: k = 9.0×10⁹ Nm²/C².

Substituting into equation 2

r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)

r = √(14364×10⁻³/5.05)

r = √(14.364/5.05)

r = √2.844

r = 1.686 m

r = 1.686 m.

Thus the distance must be 1.686 m

6 0
3 years ago
An object is dropped from 600 m. If it is initially at rest and achieves a top speed of 45 m/s just as it hits the ground, what
Brums [2.3K]

Answer:

1.68m/s^2

Explanation:

using V^2=U^2+2×a×s formula.

If,

(Final Velocity)V=45

(Initial Velocity)U=0 because it starts from rest

(Distance)S=600m

(acceleration)a= ?

now,

using formula,

45^2=0^2+2×a×600

2025= 1200a

a=2025÷1200

a = 1.68m/s^2(The unit of acceleration is m/s^2)

Therefore the acceleration is 1.68m/s^2

Hope it works !!!

3 0
3 years ago
Read 2 more answers
You are presented with several wires made of the same conducting material. The radius and drift speed are given for each wire in
jekas [21]

\begin{array}{ccc}&\text{Radius} & \text{Drift Speed}\\d) & 2 \; r &2.5 \; v\\a) & 3 \; r &1 \; v\\b) & 4 \; r &0.5 \; v\\c) & 1 \; r &5 \; v\\\end{array}

<h3>Explanation</h3>

I = v \cdot A \cdot n \cdot q,

where

  • I is the current;
  • v is the drift speed;
  • A is the cross-section area of the wire,
  • n is the number of charge carrier per unit volume, and
  • q is the charge on each charge carrier.

Area of a circular cross-section:

A = \pi \cdot r^{2},

where

  • r is the radius of the wire.

n and q are the same for all four samples, for they are made out of the same material.

As a result, I of each wire is directly proportional to v \cdot r^{2} where the value of \pi \cdot n \cdot q is constant.

For each of the four wires:

\begin{array}{ccc|c}\\& r & v &I \propto v\cdot r^{2}\\a) & 3 & 1 & 9\\b) & 4 & 0.5 & 8\\c) & 1 & 5 & 5\\d) & 2 & 2.5 & 10\\\end{array}.

How do the four wires rank by their current?

d > a > b > c.

3 0
3 years ago
3
vovangra [49]
Your answer is. D. 10÷60 = 1.66km/hr
4 0
2 years ago
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