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2 years ago

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A person wearing a shoulder harness can survive a car crash if the acceleration is smaller than -300 m/s . assuming constant acc

eleration how far must the front end of the car collapse if 2 it crashes while going 101 km/hr?

1 answer:

mars1129 [50]2 years ago

3 0

To solve this problem, we use the equation:

<span>d = (v^2 - v0^2) / 2a</span>

where,

d = distance of collapse

v0 = initial velocity = 101 km / h = 28.06 m / s

v = final velocity = 0

a = acceleration = - 300 m / s^2

d = (-28.06 m / s)^2 / (2 * - 300 m / s^2)

<span>d = 1.31 m</span>

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What is the average power output pf a weight lifter who can lift 250 kg 2.0 m in 2.0 s

liq [111]

Answer:

Power = 2.45Kw or 2450 Watts.

Explanation:

<u>Given the following data;</u>

Mass, m = 250kg

Height, h = 2m

Time, t = 2secs

We know that acceleration due to gravity, g is equal to 9.8m/s²

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

$Power = \frac {Energy}{time}$

But Energy = mgh

Substituting into the equation, we have

$Power = \frac {mgh}{time}$

$Power = \frac {250*9.8*2}{2}$

$Power = \frac {4900}{2}$

Power = 2450 Watts

To convert to kilowatt (Kw), we would divide by 1000

Power = 2450/1000

Power = 2.45Kw.

Therefore, the average power output of the weightlifter is 2.45 Kilowatts.

8 0

2 years ago

Two tugboats are moving a barge. Tugboat A pushes on the barge with a force of 3000n. Tugboat B pulls with a force of 5000 Newto

kenny6666 [7]

The net force on the barge is 8000 N

Explanation:

In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.

In this problem, we have two forces:

The force of tugboat A, $F_A = 3000 N$ , acting in a certain direction
The force of tugboat B, $F_B = 5000 N$ , also acting in the same direction

Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get

$F=F_A+F_B = 3000 + 5000 = 8000 N$

and the direction is the same as the direction of the two forces.

Learn more about forces:

brainly.com/question/11179347

brainly.com/question/6268248

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5 0

3 years ago

The fulcrum of a first-class lever divides its 9.0 m arm into two sections—a 6.0 m arm and a 3.0 m arm. You place a rock weighin

nexus9112 [7]

For balancing the lever, force on both the sides shall be equal. so,
Force on 3 m end = m × a = 3 × 98.1 = 294.3

Now, on 6 m end, it would be: = 294.3/6 = 49.05
After rounding-off to the nearest hundredth value, it would be: 49 N

Finally, Option A would be your correct answer.

Hope this helps!

6 0

2 years ago

Kinetic friction acts on a baseball player sliding into first base. Will the player's velocity change?

Law Incorporation [45]

Yes, his velocity will decrease the further he slides.

7 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago