1 year ago

Why did the chicken cross the road

Physics

2 answers:

AleksAgata [21]1 year ago

7 0

Answer:

to get to the other side.

Explanation:

bada bam, bop, BOOM

svet-max [94.6K]1 year ago

6 0

To get to the other side

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True or false: You can calculate a three variable equation knowing only one of the variables.

makvit [3.9K]

Answer:

false

Explanation:

you cant calculate a three variable equation knowing only one of the variables

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2 years ago

Consider a horizontal, uniform board of weight 125 N and length 4 m that is supported by vertical chains at each end. A person w

Norma-Jean [14]

Answer:

b. 375 N

Explanation:

System of forces in balance

ΣFy = 0 Equation (1)

Forces acting on the board

T₁: Tension in the left chain , vertical and upward

T₂ = 250 N : Tension in the right chain , vertical and upward

W₁ = 125 N : Weight of the board , vertical and downward

W₂ = 500 N : Weight of the person , vertical and downward

Calculation of the T₁

We apply the Equation (1)

ΣFy = 0

T₁+T₂-W₁-W₂ = 0

T₁ = -T₂+W₁+W₂

T₁ = -250 N+ 125 N+ 500 N

T₁ = 375 N

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3 years ago

The potential difference between two points is 100 V. If a particle with a charge of 2 C is transported from one of these points

Aleonysh [2.5K]

Answer: 200 J

Explanation: In order to explain this we have consider that the work done in a electric field is given by:

Work= Q*ΔV=2*100=200J

4 0

2 years ago

Read 2 more answers

A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2

Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel $I=25kgm^2$

Initial angular velocity of the wheel $\omega _0=10rad/sec$

Angular acceleration $\alpha =15rad/sec^2$

(a) We know that $\omega =\omega _0+\alpha t$

We have given t = 2 sec

So $\omega =10+15\times 2=40rad/sec$

Now $\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad$

(b) After 3 sec $\omega =10+15\times 3=55rad/sec$

We know that kinetic energy is given by $Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J$

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2 years ago

The specific heat of aluminum is 0.90 J/gC . How much heat is given off when 25 grams of aluminum is cooled from 55 C to 25 C?

sp2606 [1]

Answer:

2. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/gºC? c=0.90J/g. 9 (2 sigs.)

Explanation:

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2 years ago