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2 years ago

8

Assume that on a physics diagram, you saw a croquet ball moving along the grass and above it was an arrow labeled "3.5 m/s" that

indicated BOTH the speed and direction of the ball. This is an arrow is a(n) _____.
Force arrow
Acceleration arrow
Speed arrow
Vector

1 answer:

Nady [450]2 years ago

8 0

In my opinion, the answer is D, bc a vector is <span>a quantity having direction as well as magnitude, especially as determining the position of one point in space relative to another. Therefore giving the direction and speed. hope this helps, have a nice day :)</span>

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A bus travels east for 3 km, then north for 4 km. What is its final displacement?

ikadub [295]

5 km northeast. Left and up would make northeast

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3 years ago

The angular velocity of a flywheel obeys the equa tion w(1) A Br2, where t is in seconds and A and B are con stants having numer

makkiz [27]

Answer:

$A \to rad/s$

$B \to rad/s^3$

Explanation:

$\omega_z(t)=A + Bt^2$

Required

The units of A and B

From the question, we understand that:

$\omega_z(t) \to rad/s$

This implies that each of $A$ and $Bt^2$ will have the same unit as $\omega_z(t)$

So, we have:

$A \to rad/s$

$Bt^2 \to rad/s$

The unit of t is (s); So, the expression becomes

$B * s^2 \to rad/s$

Divide both sides by $s^2$

$B \to \frac{rad/s}{s^2}$

$B \to rad/s^3$

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2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

<u>Time taken to reach 1180 m is 11.29 seconds</u>

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$

<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

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2 years ago

True or False? <br> Metrics are based on the British system of weights and measures.

Sedaia [141]

The answer I'm going with is false

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3 years ago

Read 2 more answers

A boy whirls a stone in a horizontal circle of radius 1.1 m and at height 2.1 m above ground level. The string breaks, and the s

DedPeter [7]

Answer:

$212.8 m/s^{2}$

Explanation:

Time taken by stone to cover horizontal distance

$t=\sqrt{\frac {2h}{g}}$ where t is time, h is height of whirling the stone in horizontal circle, g is gravitational constant, Substituting h for 2.1 m and g for 9.81

$t=\sqrt{\frac {2*2.1}{9.81}}$ = 0.654654 seconds

t=0.65 s

Velocity, v= distance/time

v=10/0.65= 15.27525 m/s

v=15.3 m/s

$a=\frac {v^{2}}{r}$ where r is radius of circle, substituting r with 1.1m

$a=\frac {15.3^{2}}{1.1}$

$a=212.8 m/s^{2}$

Therefore, centripetal acceleration is $212.8 m/s^{2}$

8 0

3 years ago