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3 years ago

13

How far can you get away from your little brother with a paintball marker if you can travel at 3 m/s and you have 15s before he

sees you?

1 answer:

Charra [1.4K]3 years ago

3 0

Answer:

45 meters

3 (m/s) x 15 (s) = 45 meters

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A 20kg block is sliding down a 30° ramp at a constant velocity. Calculate the normal force acting on the block. Calculate the co

tatiyna

Let me try:

Normal force= 169.74N

Coefficient of kinetic friction= 0.577

Explanation

a. For a given inclined plane, normal friction is equal to the force perpendicular to the plane which is equal to

mgcos theta = 20×9.8× cos30 = 169.74N

b. The coefficient of kinetic friction for an inclined plane is given as

tan theta =tan 30 = 0.577

8 0

3 years ago

On a hot summer day, a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial veloc

hodyreva [135]

Answer:

5.52 m

Explanation:

Using the equation;

Vy = Vo sin θ

Where; Vo = 2.25 m/s and θ = 35°

We get;

= 2.25 sin 35°

= 1.29 m/s

Thus; Vy = 1.29 m/s

But a = -9.81 m/s (against gravity)

t = 1.20 s and

y = Vyt + 1/2 at²

where y is the height above the water

= 1.29 × 1.2 + (1/2 × -9.81 × 1.2²)

= 1.548 + (-7.0632)

= - 5.5152 m

Thus; the height of the girl above the water is 5.52 m

4 0

3 years ago

notka56 [123]

Answer:

toaster -- 15 A, 8 Ω
fry pan -- 10.83 A, 11.08 Ω
lamp -- 0.83 A, 144 Ω
fuse will blow

Explanation:

P = VI

I = P/V = P/120

R = V/I = V/(P/V) = V^2/P = 14400/P

<u>Toaster</u>: I = 1800/120 = 15 . . . amps

R = 14400/1800 = 8 . . . ohms

<u>Fry pan</u>: I = 1300/120 = 10.833 . . . amps

R = 14400/1300 = 11.08 . . . ohms

<u>Lamp</u>: I = 100/120 = 0.833 . . . amps

R = 14400/100 = 144 . . . ohms

The total current exceeds 20 A, so will blow the fuse.

5 0

4 years ago

In a series circuit, each circuit element has the same:a.currentc.capacitanceb.voltaged.resistance

VARVARA [1.3K]

In a series circuit, each circuit element has the same current. The answer is letter A. This is because the electrons flowing in a current has only a single path to follow. There are no other path that the current moves along.

8 0

4 years ago

A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m

Gekata [30.6K]

Answer:

819.78 m

Explanation:

<u>Given:</u>

OA = range of initial position of the airplane from the point of observation = 375 m
OB = range of the final position of the airplane from the point of observation = 797 m
$\theta$ = angle of the initial position vector from the observation point = $43^\circ$
$\alpha$ = angle of the final position vector from the observation point = $123^\circ$
$\vec{AB}$ = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

$\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m$

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

$\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} = (-708.34\ \hat{i}+412.67\ \hat{j})\ m$

The magnitude of the displacement vector = $\sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m$

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

8 0

3 years ago