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Pepsi [2]
2 years ago
5

Answer the following questions.

Physics
1 answer:
OverLord2011 [107]2 years ago
8 0
For question one, you look at the specific heats of both land and water, the water has a higher specific heat, therefore, it takes more energy in order to raise it up another degree of Celsius. What specific heat also tells us is that it takes longer for the water to cool down, therefore, land heats up more during the day, but water will cool down slower at night.
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During a field experiment about speed, a scientist created the chart above. The chart shows distance and time measurements for a
BaLLatris [955]

Answer:

yes

Explanation:

its not a good thing for the rest of your life but you have a PS4

6 0
2 years ago
A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial
Nataly_w [17]

Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

qvB=m\frac{v^2}{r}

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

r=\frac{mv}{qB}

For the ion of oxygen-16, we have:

m_A=2.66\cdot 10^{-26}kg

q_A = 1.6\cdot 10^{-19}C (it is singly charged)

v_A=2.90\cdot 10^6 m/s

B_A=1.30 T

So the radius of its orbit is

r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m

For the ion of oxygen-18, we have:

m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg

q_B = 1.6\cdot 10^{-19}C (it is singly charged)

v_B=2.90\cdot 10^6 m/s

B_B=1.30 T

So the radius of its orbit is

r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

d=2(r_B-r_A)=2(0.417-0.371)=0.092 m

3 0
3 years ago
A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential ene
Yuri [45]

Explanation:

The given data is as follows.

            U_{a} = 5.4 \times 10^{-8} J

        W_{/text{a to b}} = -1.9 \times 10^{-8} J

        Electric potential energy (U_{b}) = ?

Formula to calculate electric potential energy is as follows.

            U_{b} = U_{a} - W_{/text{a to b}}

                        = 5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)

                        = 7.3 \times 10^{-8} J

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is 7.3 \times 10^{-8} J.

6 0
3 years ago
A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.64 10-2 kg/s. the density of the gasoline is 735
Irina18 [472]
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m

Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²

Let v =  speed of pumping the gasoline, m/s
Then the mass flow rate is 
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s

The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s

Answer:  2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
7 0
3 years ago
Read 2 more answers
The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a
mario62 [17]

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

3 0
3 years ago
Read 2 more answers
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