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3 years ago

Hurry help please, number 3

Physics

2 answers:

STatiana [176]3 years ago

8 0

C) Most of earths weather occurs in this layer.

Dovator [93]3 years ago

6 0

C most of the Earths weather occurs in this layer.

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AJUDEM ME POR FAVOR URGENTE! :(

loris [4]

Link provided in other answer is a SCAM, don’t click on the link !!!!

5 0

3 years ago

A motorbike accelerates from 15m/s to 25m/s in 15 seconds.

RSB [31]

distance traveled by a uniformly accelerated bike is given as

$d = \frac{v_f + v_i}{2} (t)$

here we know that

$v_f = 25 m/s$

$v_i = 15 m/s$

$t = 15 s$

now we will have from above equation

$d = \frac{15 + 25}{2} (15)$

$d = 20 (15) = 300 m$

so it will cover the total distance of 300 m

5 0

4 years ago

Five-gram samples of copper and aluminum are at room temperature. Both receive equal amounts of energy due to heat flow. The spe

laila [671]

m₁ = mass of sample of copper = m₂ = mass of sample of aluminum = 5 g

T = initial temperature of copper = initial temperature of aluminum

T₁ = final temperature of copper

T₂ = final temperature of aluminum

c₁ = specific heat of copper = 0.09 cal/g°C

c₂ = specific heat of aluminum = 0.22 cal/g°C

Since both receive same amount of heat, hence

Q₁ = Q₂

m₁ c₁ (T₁ - T) = m₂ c₂ (T₂ - T)

(5) (0.09) (T₁ - T) = (5) (0.22) (T₂ - T)

T₁ - T = (2.44) (T₂ - T)

Change in temperature of copper = (2.44) change in temperature of aluminum

hence the correct choice is

c. The copper will get hotter than the aluminum.

4 0

3 years ago

Read 2 more answers

Two rockets are fired at each other with initial velocities of 150m/s150m/s and are 6000m6000m apart. The first rocket is accele

Nataliya [291]

Answer:

3469.788 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

First rocket

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 5\times t^2\\\Rightarrow s=150t+2.5t^2\ m$

Second rocket

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 15\times t^2\\\Rightarrow s=150t+7.5t^2\ m$

When this will collide the total distance they would have covered would be 6000 m.

$6000=150t+2.5t^2+150t+7.5t^2\\\Rightarrow 6000=300t+10t^2\\\Rightarrow 10t^2+300t-6000=0$

$t=5\left(\sqrt{33}-3\right),\:t=-5\left(3+\sqrt{33}\right)\\\Rightarrow t=13.72, -43.72$

Hence at 13.72 seconds they will collide assuming they are launched at the same time.

$s=150t+7.5t^2\\\Rightarrow s=150\times 13.72+7.5\times 13.72^2\\\Rightarrow s=3469.788\ m$

The second rocket would have gone 3469.788 m when they collide

7 0

4 years ago

A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer t

vlada-n [284]

Answer:

The correct answer is " $4.49\times 10^{10} \ joules$ ".

Explanation:

According to the question,

The work will be:

⇒ $Work=-\frac{kQq}{R}$

$=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}$

$=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}$

$=\frac{0.3978}{\varepsilon }$

$=4.49\times 10^{10} \ joules$

Thus the above is the correct answer.

3 0

3 years ago