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frez [133]
3 years ago
5

An unknown mass of each of the following substances, initially at 23.0, absorbs 1910 of heat. The final temperature is recorded

as indicated. Find the mass of the substance, given: Pyrex glass (55.5)
Physics
1 answer:
babymother [125]3 years ago
3 0
The data not given here is the specific heat capacity of pyrex glass equal to 0.75(J/g  °C). In this case, heat is equal to mass x specific heat capacity x temperature rise. Temperature difference is then equal to 1910 J / 0.75 J/g  °C / 23 g equal to 110. 75  <span>°C </span>
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A body 'A' of mass 1.5kg travelling along the positive X-axis with speed of 4.5m/s collides with another body 'B' of mass 3.2kg,
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The part of an atom that is mostly empty space is the
Ghella [55]
<h3><u>Answer;</u></h3>

Electron cloud

<h3><u>Explanation;</u></h3>
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5 0
3 years ago
Read 2 more answers
A car is moving at a rate of 72 km/hr .How far does car move when it stop after 4 seconds? ​
Zanzabum

Answer:

Assuming it starts at 72 kmph and hits a dead stop: Divide 72 by 60 for distance per minute. So, 1.2km per minute. 1.2km is 1200m and 4 seconds is one fifteenth of a minute.

Explanation:

5 0
3 years ago
If an impulse of 400 Ns acts on an object for 15s, what is the force of the object?
Ksju [112]

Answer:

J for impulse

t for time

F for force

formula is J=F×t

Explanation:

putting values in eqs after rearranging

we need to find force so

F=J ÷t

F=400÷15

=26.67

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27N is the Force applied.

7 0
3 years ago
A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130
konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

W=Fd      (1)

F: applied force = 130N

d: distance = 5.0m

W=(130N)(5.0m)=650J

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

W = 0J

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

\Delta K=W_T         (3)

You calculate the total work:

W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

Next, you replace the values of all parameters in the equation (4):

W_T=(130N-105.84N)(5.00m)=120.80J

\Delta K=120.80J

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

W_T=\frac{1}{2}M(v_2^2-v_1^2)       (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}

The final speed of the box is 2.58m/s

5 0
3 years ago
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