Which of the following is not true about tectonic plates

A body A of mass 1.5kg, travelling along the positive x-axis with speed 4.5m/s, collides with another body B of mass 3.2kg which

Veronika [31]

We consider the momentum in the x-direction and apply the principle of conservation of momentum to form the equation:
m(A)u(A) = m(A)v(A) + m(B)v(B), since u(B) = 0 as B is at rest

We calculate v(A) using:
Vx = Vcos∅
Vx = 2.1cos(30)
Vx = 1.82 m/s

1.5 x 4.5 = 1.5 x 1.82 + 3.2v(B)
v(B) = 1.26 m/s

The deflection angle of B will be 30° above the positive x-axis, so:
v(B) = Vcos∅
V = 1.26 / cos(30)
V = 1.45 m/s

The velocity of B is 1.45 m/s

7 0

4 years ago

A cube has a density of 1800 kg/m3 while at rest in the laboratory. What is the cube's density as measured by an experimenter in

olga2289 [7]

Answer:

4341.44763 kg/m³

Explanation:

$\rho'$ = Actual density of cube = 1800 kg/m³

$\rho$ = Density change due to motion

v = Velocity of cube = 0.91c

c = Speed of light = $3\times 10^8\ m/s$

Relativistic density is given by

$\rho=\frac{\rho'}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow \rho=\frac{1800}{\sqrt{1-\frac{0.91^2c^2}{c^2}}}\\\Rightarrow \rho=\frac{1800}{\sqrt{1-0.91^2}}\\\Rightarrow \rho=4341.44763\ kg/m^3$

The cube's density as measured by an experimenter in the laboratory is 4341.44763 kg/m³

5 0

4 years ago

What is the pressure inside such a balloon (in atm) if it starts out at sea level with a temperature of 12.8°C and rises to an a

tatiyna

Answer:

The final pressure inside the balloon will be $P_2=0.0359\ atm$ .

Explanation:

Given initial temperature of the balloon is $T_1=12.8\°C$

And the final temperature of the balloon is $T_2=-47.7\°C$

Initially, the balloon is at atmospheric pressure $P_1=1\ atm$ and volume be $V_1$

And the final pressure in the balloon is $P_2$ and the volume be $V_2$

Also, it is given in the question that the final volume became twenty-two times the original volume.

We can write $V_2=22V_1$

Now, using ideal gas equation.

$\frac{P_2V_2}{P_1V_1}=\frac{nRT_2}{nRT_2}$

Where, $R$ is the gas constant. And $n$ is moles of substance inside the balloon.

In our problem the value of $n$ will be the same for both cases. Also, $R$ is the gas constant.

$\frac{P_2V_2}{P_1V_1}=\frac{T_2}{T_1}$

Also, we have $V_2=22V_1$ from the question.

$\frac{P_2\times 22V_1}{P_1V_1}=\frac{T_2}{T_1}\\\\\frac{P_2\times 22}{P_1}=\frac{T_2}{T_1}$

We need to convert the temperature from $\°C$ to Kelvin.

$T_1=12.8\°C\\T_1=12.8+273.15=285.95\\T_2=-47.7\°C\\T_2=-47.7+273.15=225.45$

Plug these values we get,

$\frac{P_2\times 22}{1}=\frac{225.45}{285.95}\\\\22\times P_2=0.789\\P_2=\frac{0.789}{22}\\P_2=0.0359\ atm.$

So, the final pressure inside the balloon will be $P_2=0.0359\ atm$ .

6 0

3 years ago

Tarzan, who weighs 825 N, swings from a cliff at the end of a 19.7 m vine that hangs from a high tree limb and initially makes a

kodGreya [7K]

Answer:

a) T = (281.47 i ^ + 714.56 j ^) N , b) F_net = (281.47 i ^ - 110.44 j ^) N ,

c) F = 281.70 N, d) θ = 338.58º , e) a = 3,588 m / s² , f) θ = 201.45º

Explanation:

For this exercise we will use Newton's second law on each axis

X axis

-Tₓ = m aₓ

Y Axisy

$T_{y}$ –W = m a_{y}

Let's use trigonometry to find the components of force

sin 21.5 = Tₓ / T

cos 21.5 = T_{y} / T

Tₓ = T sin 21.5

T_{y} = T cos 21.5

Tₓ = 768 sin 21.5 = 281.47 N

T_{y} = 768 cos 21.5 = 714.56 N

a) the force of the rope on Tarzan is

T = (281.47 i ^ + 714.56 j ^) N

b) The net force is the subtraction of the tension minus the weight of Tarzan

Y Axis F_net = 714.56 - 825 = -110.44 N

F_net = (281.47 i ^ - 110.44 j ^) N

c) Let's use Pythagoras' theorem

F = √ (Fₓ² + T_{y}²)

F = √ (281.47² + 110.44²)

F = 281.70 N

d) Let's use trigonometry

tan θ = F_{y} / Fₓ

θ = tan⁻¹ F_{y} / Fₓ

θ = tan⁻¹ (-110.44 / 281.47)

θ = -21.42º

This angle is average clockwise, for counterclockwise measurement

θ = 360 - 21.42

θ = 338.58º

Acceleration

X axis

Tₓ = m aₓ

aₓ = Tₓ / m

The mass of Tarzan is

m = W / g

m = 825 / 9.8 = 84.18 kg

aₓ = 281.47 / 84.18

aₓ = -3.34 m / s2

Y Axis

T_{y}-W = m a_{y}

a_{y} = (T_{y} -W) / m

a_{y} = (714.56-825) / 84.18

a_{y} = - 1,312 m / s²

Acceleration Module

a = √ aₓ² + a_{y}²

a = √ (3.34² +1.312²)

a = 3,588 m / s²

The angle

θ = tan⁻¹ a_{y} / aₓ

θ = tan⁻¹ (-1312 / -3.34)

θ = 21.45º

Notice that the two components of the acceleration are negative, so the angle is in the third quadrant, to measure from the x-axis

θ = 180 + 21.45

θ = 201.45º

3 0

3 years ago

Summarize what you learned this week about the electromagnetic spectrum

svetoff [14.1K]

Answer:

I only really know the "How do we use these EM waves in our lives?" part srry

Explanation:

EM waves are used to make sure you cellphone, radio, TV, and etc. have service/ connection.

8 0

3 years ago

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Which of the following is not true about tectonic plates￼

Which of the following is not true about tectonic plates