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2 years ago

A 1,500 kg car hits a boulder with a force of 500 N, and the collision takes 0.5

Physics

1 answer:

Anuta_ua [19.1K]2 years ago

7 0

Answer: C.

250 kg-m/s

Explanation:

Given that the

Mass M = 1,500 kg

Force F = 500 N

Time t = 0.5 seconds

From Newton's second law of motion which state that the rate of change of momentum is proportional to the applied force.

F = mV/t

Ft = mV

Where ft = impulse: the product of time and applied force

Substitutes force and time into the formula

Ft = 500 × 0.5 = 250 Ns

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Answer:

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1 year ago

A subatomic particle created in an experiment exists in a certain state for a time of before decaying into other particles. Appl

dem82 [27]

Answer:

Δm Δt> h ’/ 2c²

Explanation:

Heisenberg uncertainty principle, stable uncertainty of energy and time, with the expressions

ΔE Δt> h ’/ 2

h’= h / 2π

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E = m c²

let's replace

Δ (mc²) Δt> h '/ 2

the speed of light is a constant that we can condense exact, so

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4 0

2 years ago

A pilot wants to fly on a bearing of 74.9 degrees. By flying due east, he find s that a 42-mph wind, blowin from the south, puts

hichkok12 [17]

The bearing shows the angle from north to the line (c), you want to see the angle inside the triangle so that will be 90-[bearing].

a=42 (42 mph wind blowing north) 

A= 90-74.9 
A= 15.1 degrees 

Ground speed is the speed the plane is going including the wind. 
ground speed = c 
Airspeed = b 

You have the angle, and you have the "Opposite" and want to find the "Hypotenuse". 
SOH CAH TOA 

Sin x = Opp/Hypot 

Sin(15.1) = 42/c 
c = 42/Sin(15.1) 
c = 161.53 

Ground speed of the plane is 161.53 mph 

Airspeed = b 

Tan x = Opp/Adj 
Tan (15.1) = 42/b 
b = 42/Tan(15.1) 
b = 161.53 

Airspeed = 161.53 mph 

(so the answer is ground speed of 161.53mph)

3 0

2 years ago

In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha

Assoli18 [71]

Answer:

$\mathbf{v_a = 4.06 \times 10^7 \ m/s}$

Explanation:

$\text{The missing part of the question is attached below.} \\ \\ \text{ mass m = 4u and charge q = 2e. Suppose a uranium nucleus with 92 protons decays into }$ $\text{into thorium, with 90 protons, and an alpha particle. The alpha particle is initally at rest at }$ $\text{ the surface of the thorium nucleus, which is 15 fm in diameter. What is the speed of the alpha}$ $\text{particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest. }$

$\text{So, from the above infromation;}$

$radius (r) = \dfrac{15 \ fm }{2}$

$radius (r) = 7.5 \times 10^{-15} \ m$

$\text{Using the formula for potential energy}$

$U = \dfrac{(9\times 10^9 \ Nm^2/C^2)(2(1.6\times10^{-19} \ C ))(90)(1.6\times 10^{-19} C)}{7.5 \times 10^{-13} \ m}$

$U = 5.529 \times 10^{-12} \ J$

$\text{Now, the speed of the alpha particle can be estimated from the conservation of energy principle}$ $\dfrac{1}{2}mv_a^2= 5.529 \times 10^{-12} J$

$v_a = \sqrt{\dfrac{2(5.529 \times 10^{-12} \ J)}{4(1.67 \times 106{-37} \ kg)}}$

$\mathbf{v_a = 4.06 \times 10^7 \ m/s}$

4 0

2 years ago

Suppose you performed the experiment in atmosphere of Argon at 25 deg. C, (viscosity of argon is 2.26X10^-5 N.s/m^2 at that temp

yuradex [85]

Answer:

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Explanation:

Remember that the net force will be zero at terminal voltege so

Mg = 6πrng

At 35v

We have

qvr = 6πrng

q= 6 x 3.142* nx 2.6*10^-5/35

q,= 3.2x 10^ - 10C

So using n= q/e

= 3.2x 10^ - 10C/1.6*10-19

= 2*10^9electrons

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2 years ago