Energy that does not involve the large-scale motion or position of objects in a system is called:

A truck covers 40.0 m in 7.50 s while uniformly slowing down to a final velocity of 2.55 m/s. (a) Find the truck's original spee

iris [78.8K]

Answer:

(a) 8.117 (b) 0.742 $m/sec^2$

Explanation:

We have given distance s =40 m time t=7.5 sec

Final velocity v =2.55 m/sec

From the first equation of motion $v=u-at$ (negative sign because there is retrdation as the truck speed is slowing down )

So $2.55=u-7.5a$ --------------eqn 1

From the second equation of motion $s=ut-\frac{1}{2}at^2$ ( negative sign because there is retrdation as the truck speed is slowing down )

So $40=7.5u-0.5\times a\times 7.5^2$

$40=7.5u-28.125a$ ------------------eqn 2

On solving eqn1 and eqn 2

u=8.117 m/sec and a=-0.742 $m/sec^2$

7 0

3 years ago

A 2kg watermelon is dropped from a 4m tall roof a) use the appropriate kinematic equations to determine the instantaneous veloci

Ivenika [448]

Answer:

8.85m/s

Explanation:

The potential energy the watermelon held before dropping is Ep=mgh=2*9.8*4=78.4J.

When it strikes the ground, all of its Ep will transfer into Ek, so 1/2*m*v^2=78.4.

We already knew that m=2, so insert that in, we will get the V^2=78.4 m/s, V=8.85 m/s

6 0

3 years ago

Which was the first object made by humans to orbit earth?

Romashka-Z-Leto [24]

The first human made object made to orbit earth was the Soviet Union's Sputnik 1, it launched the 4th of October 1957.

6 0

3 years ago

A bicyclist moves along a straight line with an initial velocity vo and slows downs. Which of the following the best describes t

xeze [42]

????????????????????

3 0

3 years ago

Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?

liberstina [14]

Answer : The maximum concentration of silver ion is $5\times 10^{-12}m$

Solution : Given,

$K_{sp}$ for AgBr = $5\times 10^{-13}$

Concentration of NaBr solution = 0.1 m

The equilibrium reaction for NaBr solution is,

$NaBr(aq)\rightleftharpoons Na^++Br^-$

The concentration of NaBr solution is 0.1 m that means,

$[Na^+]=[Br^-]=0.1m$

The equilibrium reaction for AgBr is,

$AgBr\rightleftharpoons Ag^++Br^-$

At equilibrium s s

The expression for solubility product constant for AgBr is,

$K_{sp}=[Ag^+][Br^-]$

The concentration of $Ag^+$ = s

The concentration of $Br^-$ = 0.1 + s

Now put all the given values in $K_{sp}$ expression, we get

$5\times 10^{-13}=(s)(0.1+s)$

By rearranging the terms, we get the value of 's'

$s=5\times 10^{-12}m$

Therefore, the maximum concentration of silver ion is $5\times 10^{-12}m$ .

4 0

3 years ago

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