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kotegsom [21]
4 years ago
13

A rock is thrown horizontally from a bridge with a velocity of 17.0 m/s. It takes the rock 3.0 s to strike the water below. What

is the magnitude of the velocity of the rock just before it hits the water
Physics
1 answer:
zalisa [80]4 years ago
7 0

Answer:

V= 33.98 m/s

Explanation:

Given that

Horizontal speed ,u= 17 m/s

Time taken by rockets to strike the water ,t= 3 s

We know that acceleration due to gravity ,g= 9.81 m/s²

There is no any acceleration in the horizontal direction that is why the horizontal veloity will remain constant.

In the vertical direction

vy = uy+ g t

Initial velocity in vertical direction is 0 m/s.

vy= 0+ 9.81 x 3

vy = 29.43 m/s

The resultant velocity

V=\sqrt{v_y^2+u^2}

V=\sqrt{29.43^2+17^2}\ m/s

V= 33.98 m/s

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The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 ​5 ​​ m/s then ho
Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10​⁵​​ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

F = \frac{MV}{t}

solving this two equations together;

\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0
3 years ago
The driver of a car travels at 90 km / h, observes some children playing on the road 50 m away, and applies the brakes, managing
Orlov [11]

Answer:

13,750 N

Yes

Explanation:

Given:

v₀ = 90 km/h = 25 m/s

v = 0 m/s

t = 4 s

Find: a and Δx

a = Δv / Δt

a = (0 m/s − 25 m/s) / (4 s)

a = -6.25 m/s²

F = ma

F = (2200 kg) (-6.25 m/s²)

F = -13,750 N

Δx = ½ (v + v₀) t

Δx = ½ (0 m/s + 25 m/s) (4 s)

Δx = 50 m

6 0
3 years ago
An object pulled to the right by two forces has an acceleration of 2.5m/s2. The free-body diagram shows the forces acting on the
Vanyuwa [196]

Answer:

490 N is the correct answer.

Explanation:

8 0
3 years ago
If you're going 80 mph how long does it take to go 80 miles
Elza [17]
1 mile. Is this a joke lol
6 0
3 years ago
A lacrosse ball that is thrown straight upwards reaches a maximum height of 4.5 m. At what initial velocity was it thrown? (note
shtirl [24]

Answer:

The initial velocity was 9.39 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

The ball is thrown straight upward with initial velocity u

The ball reaches a maximum height of 4.5 m

At the maximum height velocity v = 0

The acceleration of gravity is -9.8 m/s²

We need to find the initial velocity

The best rule to find the initial velocity is <em>v² = u² + 2ah</em>, where v is

the final velocity, u is the initial velocity, a is the acceleration of

gravity and h is the height

⇒ v = 0 , h = 4.5 m , a = -9.8 m/s²

⇒ 0 = u² + 2(-9.8)(4.5)

⇒ 0 = u² - 88.2

Add 88.2 to both sides

⇒ 88.2 = u²

Take square root for both sides

⇒ u = 9.39 m/s

<em>The initial velocity was 9.39 m/s</em>

5 0
3 years ago
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