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2 years ago

5

A rock climber weighs 550N and is carrying a backpack that weighs 50N. How much total work has been done after she climbs 12M

1 answer:

kherson [118]2 years ago

4 0

Answer:

7200 J

Explanation:

Work = force × distance

W = (550 N + 50 N) × (12 m)

W = 7200 J

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Answer these, and tell me HOW you got to the answer.

alex41 [277]

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4 0

2 years ago

A box slides with uniform acceleration up an incline. The box has an initial speed of 9.0 m/s and rises vertically 2.60 m before

Tju [1.3M]

Answer: 0.58

Explanation:

First we need to get the acceleration of the body using equation of motion

v²=u²-2as

v is the final velocity

u is the initial velocity

a is acceleration

s is the distance moved

0²=9²-2a(2.6)

-81=-5.2a

a=81/5.2

a= 15.6m/s²

Angle of inclination =30°

To get the coefficient of friction, we use the formula

Ff =nR

Ff is frictional force

n is coefficient of friction

R is normal reaction

n = Ff/R = Wsin30°/Wcos30°

n = tan30°

n = 0.58

4 0

2 years ago

You flip a coin straight up if the coin reaches a high point of 0.25 m above where you released it what was the initial speed?

Julli [10]

Answer: 2.21 m/s

Explanation:

Given

Coin reaches a height of 0.25 m above the launch point

Suppose u is the initial speed of the coin

Using the equation of motion

$\Rightarrow v^2-u^2=2as\\$

Putting values

$\Rightarrow 0^2-u^2=2\times g\times s=2\times (-9.8)\times 0.25\\\Rightarrow u^2=4.9\\\Rightarrow u=2.21\ m/s$

Thus, the initial velocity is 2.21 m/s

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2 years ago

What element shares the property of forming ions with a 1+ charge?

Oduvanchick [21]

Answer:

the Group 1A metals such as sodium and potassium form +1 charges,

Explanation:

5 0

2 years ago

Read 2 more answers

a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet

ANTONII [103]

Answer:

Explanation:

Given

radius of circular path $r=4\ in.$

Position is given by

$\theta =\cos 2t---1$

Differentiate 1 to angular velocity we get

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2$

Differentiate 2 to get angular acceleration

$\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3$

Net acceleration is the vector summation of tangential and centripetal force

$a_t=\alpha \times r$

$a_t=-4\cos 2t\times 4=-16\cos 2t$

$a_r=\omega ^2\cdot r$

$a_r=(-2\sin 2t)^2\cdot 4$

$a_r=16\sin^2(2t)$

$a_{net}=\sqrt{a_r^2+a_t^2}$

$a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}$

$a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}$

6 0

2 years ago