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3 years ago

6

A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary

1 answer:

Archy [21]3 years ago

5 0

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

m = 10 kg
g = 9.8 m/s²
T = ?

Use formula:

$\boxed{\bold{T=m*g}}$

Replace and solve:

$\boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}$

$\boxed{\boxed{\bold{T=98\ N}}}$

The tension in the rope is <u>98 Newtons.</u>

Greetings.

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A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w

inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is $v=\frac{d}{t}$ , where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is $d=v\cdot t$

Now, the distance equation for the man would be:

$d_{man}=v_{man}\cdot t=1.6\cdot t$

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

$d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)$

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

$d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0$

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

$d_{man}=d_{dog}$

$1.6\cdot t=3\cdot (t-1.8)$

$1.6\cdot t=3\cdot t-5.4$

$1.4\cdot t=5.4$

$t=\frac{5.4}{1.4}$

$t=3.8571s$

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

$d_{man}=1.6\cdot t=1.6\cdot (3.8571)$

$d_{man}=6.1714m$

Finally, the dog catches up with the man 6.1714m later.

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3 years ago

A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ign

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B- the acceleration is greater for the more massive rock

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2 years ago

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3 years ago

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1) Think of planting a tree. You dig the hole, put it in, and fill it back in. Next,

pogonyaev

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2 years ago

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.4 kg gibbon

Katarina [22]

Answer:

upward force acting = 261.6 N

Explanation:

given,

mass of gibbon = 9.4 kg

arm length = 0.6 m

speed of the swing

net force must provide

$F_{branch} + F_{gravity}=F_{centripetal}$

force of gravity = - mg

$F_{branch}=F_{centripetal}-F_{gravity}$

= $\dfrac{mv^2}{r} + mg$

= $m(\dfrac{3.4^2}{0.6} +9.8)$

=9 x 29.067

= 261.6 N

upward force acting = 261.6 N

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3 years ago