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3 years ago

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tyhe mass of an object is 117 g adding 1200j of heat will raise the temperture of the object by 12 celsius what is the specifc h

eat of the object A 0.35 j/g *k B 6.8 j/g *k C 0.85 j/g*k D 42 j/g*k

1 answer:

Sati [7]3 years ago

8 0

Answer:

C 0.85 j/g*k

Explanation:

The specific heat capacity of a material is given by:

$C_s = \frac{Q}{m \Delta T}$

where

Q is the amount of heat supplied to the object

m is the mass of the object

$\Delta T$ is the increase in temperature of the object

For the object in this problem, we have

m = 117 g is the mass

Q = 1200 J is the heat supplied

$\Delta T=12^{\circ}$ is the increase in temperature

Substituting into the formula, we find the specific heat:

$C_s = \frac{1200 J}{(117 g)(12^{\circ})}=0.85 J/gC$

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Your teacher (175kg) and the lab (15kg) are 2m apart. What is the gravitational force between them? Draw a FBD, and label

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2 years ago

Water ice has a density of 0.91 g/cm2, so it will float in liquid water. Imagine you have a cube of ice, 10 cm on a side. a. Wha

Reptile [31]

Answer:

(i) W = 8.918 N

(ii) $V = 9.1 \times 10^{-4} m^3$

(iii) d = 9.1 cm

Explanation:

Part a)

As we know that weight of cube is given as

$W = mg$

$W = \rho V g$

here we know that

$\rho = 0.91 g/cm^3$

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now the mass of the ice cube is given as

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now weight is given as

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Part b)

Weight of the liquid displaced must be equal to weight of the ice cube

Because as we know that force of buoyancy = weight of the of the liquid displaced

$W_{displaced} = 8.918 N$

So here volume displaced is given as

$\rho_{water}Vg = 8.918$

$1000(V)9.8 = 8.918$

$V = 9.1 \times 10^{-4} m^3$

Part c)

Let the cube is submerged by distance "d" inside water

So here displaced water weight is given as

$W = \rho_{water} (L^2 d) g$

$8.918 = 1000(0.10^2 \times d) 9.8$

$d = 0.091 m$

so it is submerged by d = 9.1 cm inside water

4 0

3 years ago

ryzh [129]

Answer:

<em>The force required is 3,104 N</em>

Explanation:

<u>Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = ma

Where a is the acceleration of the object.

On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

$v_f=v_o+at$

Where:

vf is the final speed

vo is the initial speed

a is the acceleration

t is the time

Solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

We are given the initial speed as vo=20.4 m/s, the final speed as vf=0 (at rest), and the time taken to stop the car as t=7.4 s. The acceleration is:

$\displaystyle a=\frac{0-20.4}{7.4}$

$a=-2.757\ m/s^2$

The acceleration is negative because the car is braking (losing speed). Now compute the force exerted on the car of mass m=1,126 kg:

$F = 1,126\ kg * 2.757\ m/s^2$

F= 3,104 N

The force required is 3,104 N

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2 years ago