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3 years ago

To what does global warming potential GWP refer

Physics

1 answer:

Annette [7]3 years ago

8 0

GWP is how much heat a greenhouse gas traps in the atmosphere

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A particle starts with a speed of 12m/s and moves in a straight line in such a way that its speed increases by 4m/s every second

nikitadnepr [17]

Answer:

44m/s is the speed of that particle after 8 seconds

6 0

3 years ago

A 3600-N Force causes a car to accelerate at a rate of 4m/s2. What is the mass of the car?

Mademuasel [1]

Answer:

14400 kg

Explanation:

According to Newton's second law of motion, F = ma

F = 3600 N

a = 4m/s2

3600 = m/4

3600*4 = m

m = 14400 kg

Therefore the mass of the car is 14400 kg

Hope u understand

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3 0

3 years ago

A square plate of copper with 47.0 cm sides has no net charge and is placed ina region of uniform electric field of 75.0 kN/C di

timurjin [86]

Answer:

(a) Charge density σ=6.6375×10²nC/m²

(b) Total charge Q=1.47×10²nC

Explanation:

Given Data

A=47.0 cm =0.47 m

Electric field E=75.0 kN/C

To find

(a) Charge density σ

(b)Total Charge Q

Solution

For (a) charge density σ

From Gauss Law we know that

Φ=Q/ε₀.......eq(i)

Where

Φ is electric flux

Q is charge

ε₀ is permittivity of space

And from the definition of flux

Φ = EA

The flux is electric field passing perpendicularly through the surface

Put the this Φ in equation(i)

EA =Q/ε₀

where Q(charge)=σA

EA=(σA)/ε₀

E=σ/ε₀

σ=ε₀E

$=(8.85*10^{-12} )*(75.0*10^{3} )\\=6.6375*10^{-7} C/m^{2}\\=6.63*10^{2}nC/m^{2}$

σ=6.6375×10²nC/m²

For (b) total charge Q

Q=σA

$Q=(6.6375*10^{2} nC/m^{2} )(0.47m)^{2}\\ Q=1.47**10^{2}nC$

6 0

3 years ago

A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground.

Marianna [84]

Answer:

Part a)

Final speed of the corn is 19.05 m/s

Part b)

Kinetic energy of the corn is 3.1 J

Explanation:

Part a)

As we know that the initial position of the corn is

h = 18.5 m

now we also know that it will fall from rest and moving under constant acceleration so we will have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(9.81)(18.5)$

$v_f = 19.05 m/s$

Part b)

Kinetic energy of the corn is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.017)(19.05)^2$

$K = 3.1 J$

4 0

4 years ago

Dr. Kirwan is preparing a slide show that he will present to the executive board at tonight's committee meeting. He places a 3.5

lisov135 [29]

Answer:

A) d_o = 20.7 cm

B) h_i = 1.014 m

Explanation:

A) To solve this, we will use the lens equation formula;

1/f = 1/d_o + 1/d_i

Where;

f is focal Length = 20 cm = 0.2

d_o is object distance

d_i is image distance = 6m

1/0.2 = 1/d_o + 1/6

1/d_o = 1/0.2 - 1/6

1/d_o = 4.8333

d_o = 1/4.8333

d_o = 0.207 m

d_o = 20.7 cm

B) to solve this, we will use the magnification equation;

M = h_i/h_o = d_i/d_o

Where;

h_o = 3.5 cm = 0.035 m

d_i = 6 m

d_o = 20.7 cm = 0.207 m

Thus;

h_i = (6/0.207) × 0.035

h_i = 1.014 m

8 0

3 years ago