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11Alexandr11 [23.1K]
2 years ago
14

A total of 54.0 calories of heat are absorbed as 58.3 g of lead is heated from 12.0° C to 42.0° C. What is the specific heat of

lead? A) 0.029 cal gram °C B) 0.022 cal gram °C C) 0.017 cal gram °C D) 0.031 cal gram °C
Physics
2 answers:
Trava [24]2 years ago
8 0

The answer is choice D) 0.031 cal gram degrees celsuis

Yanka [14]2 years ago
5 0

Answer: D) 0.031 cal / gram °C

Explanation:

Specific Heat can be defined as the amount of heat required to raise the temperature of 1 unit of mass by 1 degree.

Q = m c ΔT

where m is the mass, c is the specific heat and ΔT is the difference in temperatures. Q = 54.0 cal

m = 58.3 g

ΔT = 42° C- 12°C = 30°C

c = Q / (mΔT)

c = 54.0 cal / (58.3 g×30°C) = 0.031 cal /gram °C

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The centripetal acceleration of an object is given by the relation,

Ac =V^2/R

where Ac = centripetal acceleration = 98 m/s^2

R = radius of rotation = 15 m

V = speed of astronaut

Hence, \frac{V^2}{15} =98

solving this we get, V = 38.34 m/s

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2 years ago
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The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p
leva [86]

1) 1.86\cdot 10^6 rad/s^2

2) 2418 rad/s

3) 27000 m/s^2

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

\theta=\omega_i t +\frac{1}{2}\alpha t^2

where:

\theta is the angular displacement

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

\theta=90^{\circ} = \frac{\pi}{2}rad is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

Solving for \alpha, we find:

\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

\omega_f = \omega_i + \alpha t

where

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

Therefore, the final angular speed is:

\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s

3)

The tangential acceleration is related to the angular acceleration by the following formula:

a_t = \alpha r

where

a_t is the tangential acceleration

\alpha is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

v=u+at

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

a=27900 m/s^2 (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

v=0+(27900)(0.0013)=36.3 m/s

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A ______ is a push or pull that one object exerts on another.
fiasKO [112]
The answer is A. force. 
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A 115-turn circular coil of radius 2.71 cm is immersed in a uniform magnetic field that is perpendicular to the plane of the coi
tester [92]

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Time taken, t = 0.133s

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Final magnetic field, Bfin = 90.5 mT = 0.0905 T

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EMF = [(Bfin - Bin) * N * A] / t

EMF = [(0.0905 - 0.0501) * 115 * pi * 0.0271²] / 0.133

EMF = (0.0404 * 115 * 3.142 * 0.0007344) / 0.133

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