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3 years ago

The temperature of the atmosphere increases with increasing altitude in the

1 answer:

5 0

<span>The troposphere gets its heat from the ground, and so temperature decreases with altitude. Warm air rises and cool air sinks and so the troposphere is unstable. In the stratosphere, temperature increases with altitude. The stratosphere contains the ozone layer, which protects the planet from the Sun's harmful UV.</span>

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A mass of 0.40 kg is suspended on a spring which then stretches 10 cm. The mass is then removed and a second mass is placed on t

Vanyuwa [196]

Answer:

x' = 1.01 m

Explanation:

given,

mass suspended on the spring, m = 0.40 Kg

stretches to distance, x = 10 cm = 0. 1 m

now,

we know

m g = k x

where k is spring constant

0.4 x 9.8 = k x 0.1

k = 39.2 N/m

now, when second mass is attached to the spring work is equal to 20 J

work done by the spring is equal to

$W = \dfrac{1}{2}kx'^2$

$20= \dfrac{1}{2}\times 39.2\times x'^2$

x'² = 1.0204

x' = 1.01 m

hence, the spring is stretched to 1.01 m from the second mass.

7 0

3 years ago

A wooden block is pulled at a constant acceleration of 1.4 m/s2. Find the net applied force on the block if its mass is 0.6 kg.

neonofarm [45]

Answer:

F = 0.84 [N]

Explanation:

In order to solve this problem we must take into account Newton's second law, which tells us that the sum of all forces applied on a body must be equal to the product of the mass of the body by its acceleration.

F = m*a

where:

F = force [N]

a = acceleration = 1.4[m/s^2]

m = mass = 0.6 [kg]

F = 0.6*1.4

F = 0.84 [N]

7 0

4 years ago

A marble runs off the edge of a table that is 1.5 m high and the marble lands 0.50 m from the base of the table. a. How much tim

Free_Kalibri [48]

Answer:

t = 0.55[sg]; v = 0.9[m/s]

Explanation:

In order to solve this problem we must establish the initial conditions with which we can work.

y = initial elevation = - 1.5 [m]

x = landing distance = 0.5 [m]

We set "y" with a negative value, as this height is below the table level.

in the following equation (vy)o is equal to zero because there is no velocity in the y component.

therefore:

$y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\ where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]$

Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.

$(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]$

3 0

4 years ago

A jet aircraft with a mass of 4,475 kg has an engine that exerts a force (thrust) equal to 60,800 N. (a) What is the jet's accel

BartSMP [9]

Answer:

a) 13.59 m/s²

b) 67.95 m/s

c) 169.875 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

m = Mass

Force

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{60800}{4475}\\\Rightarrow a=13.59\ m/s^2$

Acceleration of the jet is 13.59 m/s²

$v=u+at\\\Rightarrow v=0+13.59\times 5\\\Rightarrow v=67.95\ m/s$

Velocity attained at 5 seconds is 67.95 m/s

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{67.95^2-0^2}{2\times 13.59}\\\Rightarrow s=169.875\ m$

Distance traveled in the 5 seconds is 169.875 m

4 0

3 years ago

Read 2 more answers

A box is pulled with a horizontal force of 500N and moves 5m what is work done

dalvyx [7]

The answer to the question is shown below:

We all know that formula for solving work done is the force multiplied by distance covered:
Work done = Force x distance
Distance = 5m
Force = 500 N
Work done = 500 N * 5m
Work done = 2500 J

4 0

4 years ago