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docker41 [41]
3 years ago
9

How much heat is needed to raise the temperature of 200 g of lead (c = 0.11 kcal/kg ∙ °c) by 10 c°?

Physics
1 answer:
Rufina [12.5K]3 years ago
3 0
Heat = mass (m)*specific heat (C)* change in temperature (Δt)

In the current scenario,
mass = 200 g = 0.2 kg
C = 0.11 kCal/kg.°C
Δt = 10 °C

Therefore,
Heat = 0.2*0.11*10 = 0.22 kCal = 0.22*4186 J = 920.92 J
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What impulse will be given to an object if it experiences a force of 0.5 N for 30 seconds
morpeh [17]

Answer:

<h2>15 Ns</h2>

Explanation:

The impulse of an object can be found by using the formula

impulse = force × time

From the question we have

impulse = 0.5 × 30

We have the final answer as

<h3>15 Ns</h3>

Hope this helps you

4 0
3 years ago
The place you get your hair cut has two nearly parallel mirrors 6.5 m apart. As you sit in the chair, your head is
Ghella [55]

Complete question is;

The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?

Answer:

13 m

Explanation:

We are given;

Distance between two nearly parallel mirrors; d = 6.5 m

Distance between the face and the nearer mirror; x = 3 m

Thus, the distance between the back-head and the mirror = 6.5 - 3 = 3.5m

Now, From the given values above and using the law of reflection, we can find the distance of the first reflection of the back of the head of the person in the rear mirror.

Thus;

Distance of the first reflection of the back of the head in the rear mirror from the object head is;

y' = 2y

y' = 2 × 3.5

y' = 7

The total distance of this image from the front mirror would be calculated as;

z = y' + x

z = 7 + 3

z = 10

Finally, the second reflection of this image will be 10 meters inside in the front mirror.

Thus, the total distance of the image of the back of the head in the front mirror from the person will be:

T.D = x + z

T.D = 3 + 10

T.D = 13m

8 0
3 years ago
A small ball with mass 1.50 kg is mounted on one end of a rod 0.600 m long and of negligible mass. The system rotates in a horiz
Georgia [21]

Answer:

(A) 0.54 kg.m^{2}

(B)  0.0156 N

Explanation:

from the question you would notice that there are some missing details, using search engines you can find similar questions online here 'https://www.chegg.com/homework-help/questions-and-answers/small-ball-mass-120-kg-mounted-one-end-rod-0860-m-long-negligible-mass-system-rotates-hori-q7245149'

here is the complete question:

A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5100 rev/min. (a) Calculate the rotational inertia of the system about the axis of rotation. (b) There is an air drag of 2.60 x 10^{-2} N on the ball, directed opposite its motion. What torque must be applied to the system to keep it rotating at constant speed?.

solution

mass of the ball (m) = 1.5 kg

length of the rod (L) = 0.6 m

angular velocity (ω) = 4900 rpm

air drag (F) =  2.60 x 10^{-2} N = 0.026 N

(take note that values from the original question are used, with the exception of the air drag which was not in the original question)

(A) because the rod is mass less, the rotational inertia of the system is the rotational inertia of the rod about the other end, hence rotational inertia =mL^{2} where m = mass of ball and L =  length of rod

= 1.5 x 0.6^{2} = 0.54 kg.m^{2}

(B) The torque that must be applied to keep the ball in motion at constant speed = FLsin90

= 0.026 x 0.6 x sin 90 = 0.0156 N

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3 years ago
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I think the answer is B.
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Electronegativity? Not sure
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