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If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he

forsale [732]

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q= mC_s \Delta T$
where m is the mass of the substance, Cs is its specific heat capacity and $\Delta T$ is the increase of temperature.

If we re-arrange the formula, we get
$C_s = \frac{Q}{m \Delta T}$
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
$C_s = \frac{1495 J}{(351 g)(66.0^{\circ}C-55.0^{\circ}C)}=0.39 J/g^{\circ}C$

6 0

3 years ago

Read 2 more answers

Which of the following is not a reason fluorescent lamps are advantageous over incandescent lamps?A. Fluorescent lamps are more

mr_godi [17]

Answer:

B. Fluorescent lamps operate at a higher temperature than incandescent

Explanation:

Fluorescent lamps have a number of advantages over incandescent lamps which are given in the options given in A, C and D. The option available in B is a drawback, not an advantage. This is because it can give out and radiate more heat as a result of working at a higher temperature. Hence B option is correct.

8 0

3 years ago

7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh

pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

$F= \mu R$

where $\mu$ is the coefficient of friction = 0.02

R = Normal reaction of the load = $mgcos\theta$ = $25 \times 9.81 \times cos 10$ = $241.52N$

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

$F=0.02 \times 241.52N$

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0

3 years ago

It takes 15 min to drive 6.0 mi in a straight line to the local hospital. It takes 10 min to go the last 3.0 mi, 2.0 min to go t

Gala2k [10]

Answer:

36.87 km/h

Explanation:

Convert all the units in SI system

1 mile = 1609.34 m

d1 = 6 mi = 9656.04 m

t1 = 15 min = 15 x 60 = 900 s

d2 = 3 mi = 4828.02 m

t2 = 10 min = 10 x 60 = 600 s

d3 = 1 mi = 1609.34 m

t3 = 2 min = 2 x 60 = 120 s

d4 = 0.5 mi = 804.67 m

t4 = 0.5 min = 0.5 x 60 = 30 s

Total distance, d = d1 + d2 + d3 + d4

d = 9656.04 + 4828.02 + 1609.34 + 804.67 = 16898.07 m = 16.898 km

total time, t = t1 + t2 + t3 + t4

t = 900 + 600 + 120 + 30 = 1650 s = 0.4583 h

The ratio of the total distance covered to the total time taken is called average speed.

Average speed = 16.898 / 0.4583 = 36.87 km/h

6 0

3 years ago

A circle has an initial radius of 50ft when the radius begins decreasing at a rate of 2ft/s. what is the rate of change of the a

valkas [14]

The area of the circle with radius r is
A = πr²

The rate of change of area with respect to time is
$\frac{dA}{dt} = \frac{dA}{dr} . \frac{dr}{dt} =2 \pi r. \frac{dr}{dt}$

The rate of change of the radius is given as
$\frac{dr}{dt} =-2 \, \frac{ft}{s}$
Therefore
$\frac{dA}{dt} =-4 \pi r \, \frac{ft^{2}}{s}$

When r = 10 ft, obtain
$\frac{dA}{dt}|_{r=10 \, ft} = -40 \pi \, \frac{ft^{2}}{s}$

Answer: - 40π ft²/s (or - 127.5 ft²/s)

7 0

4 years ago