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2 years ago

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Earth's gravity acts upon objects with a steady force of __________. A. 8. 9 meters per second B. 9. 8 meters per minute C. 8. 9

meters per minute squared D. 9. 8 meters per second squared Please select the best answer from the choices provided. A B C D.

1 answer:

Mila [183]2 years ago

8 0

Answer:

9.8 meters per second squared

Explanation:

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A girl pushes a cart with a force of 10 N. if the cart has a mass of 5 kg, what is its acceleration?

nikklg [1K]

Explanation:

Force=Mass*acceleration or F=MA. To find acceleration can be rewritten as A=F/M. If you input the numbers ypu can rewrite as A= 10N/5kg, A=2

The acceleration is 2 meters per sec.

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3 years ago

Air flows through a nozzle at a steady rate. At the inlet the density is 2.21 kg/m3 and the velocity is 20 m/s. At the exit, the

Vitek1552 [10]

To solve the problem, it is necessary to apply the concepts related to the change of mass flow for both entry and exit.

The general formula is defined by

$\dot{m}=\rho A V$

Where,

$\dot{m} =$ mass flow rate

$\rho =$ Density

V = Velocity

Our values are divided by inlet(1) and outlet(2) by

$\rho_1 = 2.21kg/m^3$

$V_1 = 20m/s$

$A_1 = 60*10^{-4}m^2$

$\rho_2 = 0.762kg/m^3$

$V_2 = 160m/s$

PART A) Applying the flow equation we have to

$\dot{m} = \rho_1 A_1 V_1$

$\dot{m} = (2.21)(60*10^{-4})(20)$

$\dot{m} = 0.2652kg/s$

PART B) For the exit area we need to arrange the equation in function of Area, that is

$A_2 = \frac{\dot{m}}{\rho_2 V_2}$

$A_2 = \frac{0.2652}{(0.762)(160)}$

$A_2 = 2.175*10^{-3}m^2$

Therefore the Area at the end is $21.75cm^2$

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3 years ago

Vector c has a magnitude 24.6 m and is in the direction of the negative y-axis. vectors a and b are at angles Î± = 41.4Â° and Î²

storchak [24]

The vector c has a magnitude of 24.6m and it is in the negative y direction. Therefore
$\vec{c} = - 24.6 \hat{j}$

The vector b is 41.4° up from the x-axis. Therefore
$\vec{b} = b[cos(41.4^{o}) \hat{i} + sin(41.4^{o}) \hat{j} ] =b(0.75\hat{i} + 0.6613 \hat{j})$

The vector a is 27.7° up from the x-axis. Therefore
$\vec{a} = a[cos(22.7^{o})\hat{i} + sin(27.7^{o})\hat{j}] = a(0.8854\hat{i} + 0.4648\hat{j})$

Because $\vec{a} +\vec{b} + \vec{c} = 0$ , the sum of the x and y components should be zero. Therefore,
For the x-component,
0.8854a + 0.75b = 0
or
a + 0.847b = 0 (1)
For the y-component,
0.4648a + 0.6613b - 24.6 = 0
or
a + 1.4228b = 52.926 (2)

Subtract (1) from (2).
0.5758b = 52.926
b = 91.917
a = -0.847b = -77.854

Answer:
The magnitude of vector a is -77.85 m
The magnitude of vector b is 91.92 m

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3 years ago

The diagram shows the trajectory of a ball that

11Alexandr11 [23.1K]

Answer:

69 MPH

Explanation:

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3 years ago