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2 years ago

Please help: I don't know how to do these problems

Physics

1 answer:

antiseptic1488 [7]2 years ago

3 0

d =2.55.68m and t = 11.36s
In my opinion

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(8c7p26) During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with lar

likoan [24]

Answer: 1175 J

Explanation:

Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."

Given

Spring constant, k = 102 N/m

Extension of the hose, x = 4.8 m

from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m

Work done =

W = 1/2 k [x(i)² - x(f)²]

Since x(f) = 0, then

W = 1/2 k x(i)²

W = 1/2 * 102 * 4.8²

W = 1/2 * 102 * 23.04

W = 1/2 * 2350.08

W = 1175.04

W = 1175 J

Therefore, the hose does a work of exactly 1175 J on the balloon

7 0

2 years ago

evaluate the force of a skateboard exerts on you if your mass is 60kg and you push on the skateboard with a force of 60N

Sav [38]

Every action has an equal or opposite reaction.
You weigh 60kg
<span>So your acceleration is 6N / 60 kg = 0.1m/s^2</span>

6 0

2 years ago

39. Draw a complete free body diagram of a 40 kg plastic crate at rest on a wooden table (us=0.7). The applied force to the righ

Leya [2.2K]

In order to draw the free body diagram, first let's calculate the friction force acting on the crate:

$\begin{gathered} F_f=N\cdot\mu \\ F_f=40\cdot9.8\cdot0.7 \\ F_f=274.4\text{ N} \end{gathered}$

Since the friction force is greater than the force applied, the crate will not move, and the friction force will be equal to the force applied.

The weight force is equal to 40 * 9.8 = 392 N.

So, drawing the diagram, we have:

4 0

11 months ago

A cheetah runs with a constant speed of 8 m/s. How far has the cheetah gone after 25 seconds?

Radda [10]

Multiply 25 * 8 = 200
So the cheetah has gone 200 miles in 25 seconds

7 0

2 years ago

Read 2 more answers

A first order reaction, A -> products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,

tamaranim1 [39]

Answer: a) The rate constant, k, for this reaction is $0.00516s^{-1}$

b) No $t_{\frac{1}{2}}$ does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$A\rightarrow products$

Given: Order with respect to $A$ = 1

Thus rate law is:

a) $Rate=k[A]^1$

k= rate constant

$0.00250=k[0.484]^1$

$k=0.00516s^{-1}$

The rate constant, k, for this reaction is $0.00516s^{-1}$

b) 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 = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}$

$t_{\frac{1}{2}}=\frac{0.69}{k}$

Thus $t_{\frac{1}{2}}$ does not depend on concentration.

8 0

2 years ago