All categories

3 years ago

What are three common fossil fuels?

Physics

2 answers:

Gekata [30.6K]3 years ago

7 0

Answer:

A. coal, oil, natural gas

Explanation:

this is because those three are related to fossil fuels.

Zielflug [23.3K]3 years ago

4 0

Answer:

A:Coal,Oil,Natural gas

Explanation:

that pfp tho

You might be interested in

A rod made up of certaon metall which had coefficiebt of volume expansion is 33×10^-6 per celsius degrer.The rod is 100 meter lo

Irina-Kira [14]

Answer:

100.11 m

Explanation:

Given:

Coefficient of volume expansion, $\alpha_{v}=33\times10^{-6}$ /°C.

Initial length of rod, $l=100 m$

Change in temperature, ΔT= $(90-(-10))=100$ °C

We know that,

$\alpha_{v}=3\alpha_{L}\\\alpha_{L}=\frac{\alpha_{v}}{3}\\\alpha_{L}=\frac{33\times10^{-6}}{3}=11\times10^{-6}$ /°C

Here, $\alpha_{L}$ is the coefficient of linear expansion.

Now, we know that,

Change in length (Δl) is given as,

$\Delta l=\alpha_L\times L\times\Delta T$

Plug in all the values and solve for $l$ .

This gives,

$\Delta l=11\times10^{-6}\times 100\times100\\\Delta l=0.11 \textrm{ m}$

Therefore, the length of rod after expansion is $L+\Delta l=100+0.11=100.11$ m.

4 0

3 years ago

A 5 kg brick is dropped from a height of 12m on a spring with a spring constant 8 kN/m. If the spring has unstretched length of

Ivan

Answer:

0.1164 m

0.49387 m

Explanation:

m = Mass = 5 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height from which brick falls = 12 m

k = Spring constant = 8 kN/m

a) Potential energy of the brick

$PE=mgh\\\Rightarrow PE=5\times 9.81\times 12\\\Rightarrow PE=588.6\ J$

Potential energy in spring

$PE=\frac{1}{2}\times kx^2\\\Rightarrow x=\sqrt{\frac{PE\times 2}{k}}\\\Rightarrow x=\sqrt{\frac{588.6\times 2}{8000}}\\\Rightarrow x=0.3836\ m$

The compression of spring = 0.5-0.3836 = 0.1164 m

b) Weight of the brick

$F=5\times 9.81\\\Rightarrow F=49.05\ N$

$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{49.05}{8000}\\\Rightarrow x=0.00613\ m$

The final length of the spring = 0.5-0.00613 = 0.49387 m

4 0

3 years ago

Part B<br> Do the atoms in salt follow a pattern?

erastovalidia [21]

Answer: Yes.

Explanation: Because of the crystalline structure.

4 0

3 years ago

Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s

enot [183]

Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:

$p_i = p_f$

$m_1u_1 + m_2v_2 = (m_1 + m_2)v$

$v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}$

$v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s$

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}$

Here, initial velocity is the final velocity from the first stage. Therefore:

$v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s$

3. Use conservation of momentum to find the combined speed of Gayle and her brother.

Given:

Initial velocity of Gayle and sled is, $u_1(i)=10.5$ m/s

Initial velocity of her brother is, $u_2(i)=0$ m/s

Mass of Gayle and sled is, $m_1=55.0$ kg

Mass of her brother is, $m_2=30.0$ kg

Final combined velocity is given as:

$v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}$

$v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79$ m/s

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s$

6 0

3 years ago

2) A man squeezes a pin between his thumb and finger, as shown in Fig. 6.1.

Salsk061 [2.6K]

<h3>pressure = force / area</h3>

<h3>force = 84 N</h3><h3>pressure = 6 × 10 - 5 = 55 m2</h3>

<h3>pressure = 84 / 55</h3>

<h3>pressure = 1.53 pascals</h3>

hope that helps and please tell me if i am wrong :)

8 0

3 years ago