Ask question

Ask question

All categories

11Alexandr11 [23.1K]

3 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

2 answers:

Trava [24]3 years ago

8 0

The answer is choice D) 0.031 cal gram degrees celsuis

Yanka [14]3 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

You might be interested in

a racing car traveling initially at 8.0 m/s accelerates uniformly at 10.0 m/s^2 for 5 seconds. How far does it travel in this ti

Pepsi [2]

The car travels a distance of

(8.0 m/s) (5 s) + 1/2 (10.0 m/s²) (5 s)² = 165 m

7 0

3 years ago

Tần số của dao động duy trì

AfilCa [17]

Amman oak smoke skins kaons

8 0

3 years ago

The table shows the charges and the distance between five different pairs of objects.

zmey [24]

Answer:

the list of forces from greatest to least is Y, W, X, Z

Explanation:

Given data:

5 pairs of charged plates with force between each pair as : F, W, X,Y, Z

Also given are the charges q1 and q2 of plates in each pair and the distances between them (refer attachment)

To find: to arrange the force from greatest to least.

Initial step is to calculate the force value using the given data of charges and distance between plates.

We know that the force between two charged plates is given by the formula:

F = $\frac{kq1q2}{r^{2} }$

where,

q1 and q2 are charge of the two plates

r is the distance between two plates

k is coulomb constant

Since the problem involves comparison of forces, the k values is not necessarily discussed.

Now we have to calculate force values

F = $\frac{kq^{2} }{d^{2} }$

W = $\frac{2kq^{2} }{d^{2} }$

X = $\frac{kq^{2} }{4d^{2} }$

Y = $\frac{3kq^{2} }{d^{2} }$

Z = $\frac{kq^{2} }{9d^{2} }$

Assume values of q = 1 and d = 2, we get

F = 0.5k

W = 0.5k

X = 0.0625k

Y = 0.75k

Z =0.025k

F and W are though same, the calculated force value shows W = 2F

Hence W is greater than F.

Therefore, the list of forces from greatest to least is Y, W, F, X, Z

Hence answer is : Y, W, X, Z

7 0

3 years ago

Read 2 more answers

There is an old physics joke involving cows, and you will need to use its punchline to solve this problem

aleksandr82 [10.1K]

Answer:

The angle of deflection will be "1.07 × 10⁻⁷°".

Explanation:

The given values are:

Mass of a cow,

m = 1100 kg

Mass of bob,

mb = 1 kg

The total distance between a cow and bob will be,

d = 2 m

Let,

The tension be "t".

The angle with the verticles be " $\theta$ ".

Now,

Vertically equating forces

⇒ $T\times Cos \theta =mb\times g$ ...(equation 1)

Horizontally equating forces

⇒ $T\times Sin \theta = G\times M\times \frac{mb}{d^2}$ ...(equation 2)

From equation 1 and equation 2, we get

⇒ $tan \thata=\frac{G\times M}{g\times d^2}$

O putting the estimated values, will be

⇒ $\theta = tan(\frac{6.674\times 10^{-11}\times 1100}{9.8\times 2^2} )$

⇒ $\theta = 1.07\times 10^{-7}^{\circ}$

6 0

4 years ago

A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i

vovangra [49]

Answer:

$c=71.4m/s$

$\theta=8.54\textdegree$

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed $V_w= 15 m/s \angle 45 \textdegree$

Generally Resolving vector mathematically

$sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6$

Generally the equation Pythagoras theorem is given mathematically by

$c^2=a^2+b^2$

$c^2=10.6^2 +(10.6+60)^2$

$c=\sqrt{10.6^2 +(10.6+60)^2}$

Therefore Resultant velocity (m/s)

$c=71.4m/s$

b)Resultant direction

Generally the equation for solving Resultant direction

$\theta=tan^-1(\frac{y}{x})$

Therefore

$\theta=tan^-1(\frac{10.6}{70.6})$

$\theta=8.54\textdegree$

7 0

3 years ago