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3 years ago

A 5kg rock and a 10kg rock are dropped from a height of 10m?

1 answer:

5 0

Look out below ! You should step nimbly to one side, to avoid being hit by one or the other of those hazardous weight objects when they arrive (at the same time).

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An electronic line judge camera captures the impact of a 57.0-g tennis ball traveling at 32.2 m/s with the side line of a tennis

mote1985 [20]

Answer:

average acceleration is 1.365 × $10^{4}$ m/s²

Explanation:

given data

initila speed u = -32.2 m/s

final speed v = 21.6 m/s

time taken t = 0.00394 s

solution

we get here average acceleration that will be express as

v = u + at ..........................1

put here value and we get

21.6 = -32.2 + a × 0.00394

solve it we get

a = 1.365 × $10^{4}$ m/s²

so average acceleration is 1.365 × $10^{4}$ m/s²

3 0

3 years ago

Lab number 14: sudden stops hurt newtons first law

kompoz [17]

Answer:

you want to know newtons first law here.

Explanation:

An object that is at rest stays at rest or stays in motion.

7 0

3 years ago

The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a

bekas [8.4K]

Answer:

The work done on the canister by the 5.0 N force during this time is

54.06 Joules.

Explanation:

Let the initial kinetic energy of the canister be

KE₁ = $\frac{1}{2} mv_1^{2}$ = $\frac{1}{2} *3*3.6^{2}$ = 19.44 J in the x direction

Let the the final kinetic energy of the canister be

KE₂ = $\frac{1}{2} mv_2^{2}$ = $\frac{1}{2} *3*7.0^{2}$ = 73.5 J in the y direction

Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final

= 73.5 J - 19.44 J = 54.06 J

3 0

3 years ago

Read 2 more answers

Toy car in a science experiment covers 1.6 meters in half a second. If a the car travels at a steady speed, how far will it go i

Tanzania [10]

The answer is D. 32 m.

The simple equation that connects speed (v), time (t), and distance (d) can be expressed as:
$v= \frac{d}{t}$ ⇒ $d=v*t$

It is given:
$v = \frac{1.6m}{0.5s} = \frac{1.6m*2}{0.5s*2}= \frac{3.2m}{1s} = 3.2 m/s$
t = 10 s
d = ?

So:
$d= v*t=3.2m/s*10s = 32m$

3 0

3 years ago

A 94 g particle undergoes SHM with an amplitude of 8.3 mm, a maximum acceleration of magnitude 7.8 x 103 m/s2, and an unknown ph

Lelechka [254]

Answer:

a) T = 6.49*10^-3 s

b) v = 8 m/s

c) E = 3 J

d) F = 733 N

e) F = 366.5 J

Explanation:

Given

Mass of particle, m = 94 g = 0.094 kg

Amplitude of the particle, A = 8.3 mm = 8.3*10^-3 m

Maximum acceleration of particle, a = 7.8*10^3 m/s²

the equation describing Simple Harmonic Motion is given as

x = A cos (wt +φ)

To fond the acceleration of this relationship, we would have to integrate. Twice, the first would be a Velocity, and the second acceleration that we need.

Velocity = dx/dt = -Aw sin(wt + φ)

Acceleration = d²x/dt = -Aw² cos(wt + φ)

From the question, we were given, magnitude of acceleration to be 7.8*10^3 m/s²

Aw² = 7.8*10^3

w² = 7.8*10^3 / A

w² = 7.8*10^3 / 8.3*10^-3

w² = 939759

w = √939759

w = 969

Recall, T = 2π/w, so that

T = (2 * 3.142) / 969

T = 6.49*10^-3 s

Maximum speed = Aw

Maximum speed = 8.3*10^-3 * 969

Maximum speed = 8.0 m/s

Total mechanical energy oscillator =

mgx + 1/2mx² =

1/2mv(max)² =

1/2 * 0.094 * 8² =

3 J

Maximum displacement

x = A cos(wt + φ)

For x to be maximum here, then cos(wt + φ) Must be equal to 1

Acceleration = d²x/dt² = -Aw²

And force = mass * acceleration

Force = 0.094 * 7.8*10^3

Force = 733 N

x = A cos(wt + φ), where cos(wt + φ) = 1/2

d²x/dt² = -Aw² * 1/2

d²x/dt² = 733 * 0.5

= 366.5 N

7 0

3 years ago