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3 years ago

7

Fill in this pretty little blank.

2 answers:

IgorLugansk [536]3 years ago

6 0

Freaky girl , it from my

Trava [24]3 years ago

3 0

Answer:

freaky girl

it from my

Explanation:

lol

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⦁ Match the following terms:

ki77a [65]

Answer:

Mass number - ⦁ The number of protons and neutrons in the nucleus of an atom.

Isotopes - ⦁ Atoms with the same number of protons, but different number of neutrons.

Nitrogen - ⦁ The name of the element with atomic number 7.

Atomic number - ⦁ The number of protons in the nucleus of an atom.

4 0

3 years ago

Determine a formula for the acceleration of the system in terms of mA, mB, θ, and g. Ignore the mass of the cord and pulley. Exp

jekas [21]

Answer:

$a=\frac{mBg-mAgSin\theta}{mA+mB}$

Explanation:

Given two mass on an incline code $mA$ and $mB$ and an angle of inclination $\theta$ . $g$ . Assume that $mA$ is the weight being pulled up and $mB$ the hanging weight.

-The equations of motion from Newton's Second Law are:

$mBg-T=mBa$ where a is the acceleration.

#Substituting for $T$ (tension) gives:

$mBg-mAsin\theta-mAa=mBa$

#and solving for $a:$

$a=\frac{mBg-mAgSin\theta}{mA+mB}$ which is the system's acceleration.

8 0

4 years ago

A car moving has a speed of 35 m/s. What acceleration would it have if it took 5.0 s to come to a complete stop?

Pepsi [2]

Answer:

7

Explanation:

7 0

3 years ago

Read 2 more answers

Standing on the surface of the earth you have a weight of 100 N. If you were to travel until you were 2

Vika [28.1K]

Answer:

E

Explanation:

8 0

3 years ago

A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th

son4ous [18]

Answer:

Part a)

$v = 7.57 km/h$

Part b)

$\theta = 67.5 degree$ North of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

$d_1 = 40 km$

now it turns towards 50 degree East of North

so its distance is given as

$d_2 = 20 km(sin50 \hat i + cos50\hat j)$

$d_2 = 15.3 \hat i + 12.8 \hat j$

then finally it moves towards west for 50 min

$d_3 = -50 \hat i$

Now the total displacement of the train is given as

$d = d_1 + d_2 + d_3$

$d = (40 + 15.3 - 50)\hat i + 12.8 \hat j$

$d = 5.3\hat i + 12.8 \hat j$

now total time duration of the motion is given as

$T = 40 min + 20 min + 50 min$

$T = 1.83 h$

now average velocity is given as

$v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}$

$v_{avg} = 2.89\hat i + 6.99\hat j$

Part a)

magnitude of the average velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{2.89^2 + 6.99^2}$

$v = 7.57 km/h$

Part b)

Direction of the velocity is given as

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{6.99}{2.89}$

$\theta = 67.5 degree$ North of East

6 0

4 years ago