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4 years ago

2. If a police car drives at a constant speed of 30 km/h, how long will it take to travel a

Physics

1 answer:

Natasha_Volkova [10]4 years ago

7 0

Explanation:

Time = Distance/Speed

= 96km / (30km/h)

= 3.2h. (Or 3 hours and 12 minutes)

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A solid cylinder is released from the top of an inclined plane of height 0.81 m. From what height, in meters, on the incline sho

Jlenok [28]

Answer:

same 0.81m

Explanation:

in this problem if we assume there no resistance of any sort. and we apply the energy conservation

change in Potential energy = change in kinetic energy

mgh = 0.5mv^2

gh = 0.5v^2

the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.

so at the bottom

put h = 0.81m

9.81 * 0.81 * 2 = v^2

v=3.99 m/s

both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size

3 0

3 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!pls

barxatty [35]

I think this is the solution:

1: U-1, F,-4
2: Na-6, Mo-1, O-4
3: Bi-1, O-1, C-1, I-1
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5: N-2, H-4, S-1, C-1
6: Ge- 15, N-4
7: N-1, H-4, C-1, I-1, O-3
8: H-7, F-1
9: N-1, O-5, H-1, S-1
10: H-8
11: Nb-1, O-1, C-1, I-3
12: C-3, F-3, S-1, O-3, H-1
13: Ag-1, C-1, N-1, O-1
14: Pb-6, H-1, As-1, O-4

5 0

2 years ago

n a level football field a football is projected from ground level. It has speed 9.0 m/sm/s when it is at its maximum height. It

Pani-rosa [81]

Answer:

T = 0.225 s

Explanation:

The speed of a projectile at the highest point of its motion is the horizontal speed of the projectile. Considering the horizontal motion with negligible air resistance, we can use the following formula:

$v_x = RT\\\\T = \frac{v_x}{R}$

where,

T = Total time of ball in air = ?

R = Horizontal distance covered = 40 m

$v_x$ = horizontal speed = 9 m/s

Therefore,

$T = \frac{9\ m/s}{40\ m}$

4 0

3 years ago

Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

Step2247 [10]

Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that $F = k \frac{Q_1Q_2}{d^2}$ so if we double the charge on $Q_1$ and double the distance to $(2d)$ we plug these into the equation to find

<span> $F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}$ </span>

So we see the new force is exactly 1/2 of the old force so your answer should be $\frac{1}{2}F$ if I can remember my physics correctly.

9 0

4 years ago

Read 2 more answers

An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start

galina1969 [7]

Answer:

it moves 25 inches.

Explanation:

the east west bit isn't important, ignore it. if an ant starts at 6 then moves to 19 then we need to subtract 19 from 6, that's 13. then it moves to 7. the difference between 19 and 7 is 12. add that to 13 and you get 25. it's important to remember that there is no such thing as negative distance. if it moved, then it counts.

3 0

3 years ago