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jenyasd209 [6]
3 years ago
7

A situation that restricts something

Physics
1 answer:
Ostrovityanka [42]3 years ago
7 0
Answer:

Laws

Explanation:
i don’t have one but i hope ur doing well during this quarantine
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6. If a vehicle needs 5s to complete 15m. Find the mean speed of it?
Karolina [17]

Answer:

3m/s

Explanation:

Time=5s

Distance =15m

Speed=distance/time

Putting the values

Speed=15m/5s

Speed=3m/s is the answer

Hope it will help you. :)

4 0
3 years ago
How can an object overcome static friction?
larisa86 [58]

Answer:

Applying enough force in one direction to move the object, making kinetic energy.

Explanation:

Simpleness

4 0
3 years ago
This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a
Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = time taken for decomposition  = 3 days

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{58}{100}\times 100=58g

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

k=\frac{2.303}{3}\log\frac{100}{58}

k=0.18days^{-1}

a) Half-life of radon-222:

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant  = 0.18days^{-1}

t = time taken for decomposition  = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{15}{100}\times 100=15g

t=\frac{2.303}{0.18}\log\frac{100}{15}

t=10.54days

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0
3 years ago
Two forces, one of 100 ponds and the other 150 pounds act on the same object, at angles of 20°and 60°, respectively, withthe pos
soldi70 [24.7K]
<h2>Resultant is 235.54 pounds at an angle 44.16° to X axis.</h2>

Explanation:

Forces are 100 pound and 150 pound and angles with x axis are 20°and 60°.

That is force 1 is 100 pound with x axis at 20°

           F₁ = 100 cos 20 i  +  100 sin 20 j

           F₁ = 93.97 i  +  34.20 j          

That is force 2 is 150 pound with x axis at 60°

           F₂ = 150 cos 60 i  +  150 sin 60 j

           F₂ = 75 i  +  129.90 j  

F₁ +  F₂ =  93.97 i  +  34.20 j + 75 i  +  129.90 j

F₁ +  F₂ =  168.97 i  +  164.10 j

\texttt{Magnitude = }\sqrt{168.97^2+164.10^2}\\\\\texttt{Magnitude = }235.54pounds\\\\\texttt{Angle = }tan^{-1}\left ( \frac{164.10}{168.97}\right )\\\\\texttt{Angle = }44.16^0

Resultant is 235.54 pounds at an angle 44.16° to X axis.

6 0
3 years ago
How high would you have to lift a 1000kg car to give it a potential energy of:
Elza [17]

Given parameters:

Mass of the car = 1000kg

Unknown:

Height  = ?

To find the heights for the different amount potential energy given, we need to understand what potential energy is.

Potential energy is the energy at rest due to the position of a body.

 It is mathematically expressed as:

          P.E  = mgh

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height of the car

Now the unknown is h, height and we make it the subject of the expression to make for easy calculation.

               h = \frac{P.E }{mg}

<u>For 2.0 x 10³ J;</u>

                  h  = \frac{2000}{1000 x 9.8}   = 0.204m

<u>For 2.0 x 10⁵ J;</u>

                  h  = \frac{200000}{9.8 x 1000}   = 20.4m

<u>For 1.0kJ  = 1 x 10³J; </u>

                  h  = \frac{1000}{9.8 x 1000}   = 0.102m

   

5 0
3 years ago
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