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3 years ago

I'd like you to think back on 2 events in

Physics

1 answer:

strojnjashka [21]3 years ago

7 0

Answer:

Holi

biwali

these are best events and love to celebrate withmy family and friends it contains lot of happiness and joy

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The half-life for a 100-gram sample of radioactive element X is 5 days. How much of element X remains after 10 days have passed?

pav-90 [236]

Answer:

25g

Explanation:

Given parameters:

Half life of X = 5days

Mass of sample of radioactive elements = 100g

The half-life of a radioactive material is the time it takes for a sample to decay by half. At the half-life, an original sample would have yielded equal daughter to parent ratio.

At time (days) Mass of sample

0 100g

5 50g

10 25g

Every 5 days, the sample will split into half.

Therefore, 25g will remain after 10days.

3 0

3 years ago

All of the following except ______ were part of RCA. A. Westinghouse B. Columbia Broadcasting System C. United Fruit Company D.

PSYCHO15rus [73]

Answer:

Columbia broadcasting system was never been a part of RCA corporation. The other companies Westinghouse, united fruit company and general electric were remain as part of RCA.

Explanation:

7 0

3 years ago

A paintball is shot horizontally in the positive x direction at time t after the ball is shot it is 4 cm to the right and 4 cm b

Artist 52 [7]

<u>Answer:</u>

At time 2t the paint ball is at 8 cm to the right and 16 cm to the bottom

<u>Explanation:</u>

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Considering the horizontal motion of paint ball

Distance traveled during time t = 4 cm

Initial velocity = u m/s

Acceleration = 0 $m/s^2$

So $4 = u*t+\frac{1}{2} *0*t^2\\ \\ u = \frac{4}{t}$

Now at time 2t,

$S= u*2t+\frac{1}{2} *0*(2t)^2\\ \\=\frac{4}{t} *2t\\ \\ =8cm$

So horizontal distance traveled in time 2t = 8 cm to the right

Now considering the vertical motion of paint ball

Distance traveled during time t = 4 cm

Initial velocity = 0 m/s

Acceleration = -g $m/s^2$

$4=0*t-\frac{1}{2} *g*t^2\\ \\ t^2=\frac{-8}{g}$

At time 2t,

$S=0*2t-\frac{1}{2} *g*(2t)^2\\ \\ =-\frac{1}{2} *g*4*\frac{-8}{g}\\ \\ =16 cm$

So vertical distance traveled in time 2t = 16 cm to the bottom

6 0

3 years ago

Determine an appropriate size for a square cross-section solid steel shaft to transmit 260 hp at a speed of 550 rev/min if the m

Phantasy [73]

Answer:46.05 mm

Explanation:

Given

$Power=260\ hp\approx 260\times 746=193.96\ KW$

speed $N=550\ rev/min$

allowable shear stress $(\tau )_{max}=15\ kpsi\approx 103.421\ MPa$

Power is given by

$P=\frac{2\pi NT}{60}$

$193.96=\frac{2\pi 550\times T}{60}$

$T=3367.6\ N-m$

From Torsion Formula

$\frac{T}{J}=\frac{\tau }{r}-----1$

where J=Polar section modulus

T=Torque

$\tau$ =shear stress

For square cross section

$r=\frac{a}{2}$

where a=side of square

$J=\frac{a^4}{6}$

Substituting the values in equation 1

$\frac{3376.6}{\frac{a^4}{6}}=\frac{103.421\times 10^6}{\frac{a}{2}}$

$a=0.04605\ m$

$a=46.05\ mm$

7 0

3 years ago

What is the work done by a car's braking system when it slows the 1500-kg car from an initial speed of 96 km/h down to 56 km/h i

kondor19780726 [428]

Answer:

-352.275KJ

Explanation:

We are given that

Mass of car=1500kg

Initial speed of car =u=96 km/h= $96\times \frac{5}{18}=26.67m/s$

1km/h= $\frac{5}{18}m/s$

Final speed of car=v=56km/h= $56\times \frac{5}{18}=15.56m/s$

Distance traveled by car=s=55m

We have to find the work done by the car's braking system.

Using third equation of motion

$v^2-u^2=2as$

$(15.56)^2-(26.67)^2=2a(55)$

$-469.18=110a$

$a=\frac{-469.18}{110}=-4.27m/s^2$

Where negative sign indicates that velocity of car decreases.

Work done by a car's barking system= $w=F\times s=ma\times s$

Work done by a car's barking system= $1500\times -4.27\times 55=352275J$

Work done by a car's barking system= $-\frac{352275}{1000}=-352.275KJ$

1KJ=1000J

Where negative sign indicates that work done in opposite direction of motion.

7 0

4 years ago

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