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3 years ago

If an IPS student weighs 115 lbs on Earth, what is their weight in Newtons?

Physics

2 answers:

Montano1993 [528]3 years ago

6 0

Answer:

511.545 Newtons

Explanation:

So 1 pound=4.44822 Newton’s so 115 times 4.44822 is 511.5453, then round it to get 511.545 Newtons.

arlik [135]3 years ago

6 0

511.545 N hdjshdhejsjdjejsjjdjsjdjejjsjdjjs

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Peter’s body supplies a force of 500 N to run up a 10-m hill in 10 s. How much power is involved in Peter’s run up the hill? Exp

shutvik [7]

Answer: 500 Watts

Explanation:

Power $P$ is the speed with which work $W$ is done. Its unit is Watts ( $W$ ), being $1 W=\frac{1 Joule}{1 s}$ .

Power is mathematically expressed as:

$P=\frac{W}{t}$ (1)

Where $t$ is the time during which work $W$ is performed.

On the other hand, the Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path. It is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter ( $1J=(1N)(1m)=Nm$ ).

When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:

$W=(F)(d)$ (2)

In this case, we have the following data:

$F=500 N$

$d=10 m$

$t=10 s$

So, let's calculate the work done by Peter and then find how much power is involved:

From (2):

$W=(500 N)(10 m)$ (3)

$W=5000 J$ (4)

Substituting (4) in (1):

$P=\frac{5000 J}{10 s}$ (5)

Finally:

$P=500 W$

3 0

2 years ago

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final

Serhud [2]

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B = -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

$x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s$

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

5 0

2 years ago

State Newton’s third law of motion.

bagirrra123 [75]

Answer:every force has an equal and oppisite reaction. meeaning if you punch a wall wwith 50lbs of force the wall pushes back on your hand with 50 lbs of force

Explanation:

7 0

3 years ago

A gorilla weighs 8.00x10^2N and swings from vine to vine. As the gorilla grads a new vine, both vines make an angle of 30 degree

Yakvenalex [24]

Answer:

461.88 N

Explanation:

Let tension in each vine be T .

The vertical component of this tension of both the vine will add up and balance the weight .

2T cos 30 = mg

2 T x cos 30 = 800

T = 800 / (2 x cos30 )

= 461.88 N

6 0

2 years ago

In a wire, when elongation is 4 cm energy stored is E. if it is stretched by 4 cm, then what amount of elastic potential energy

myrzilka [38]

<h2>Answer:</h2>

<h2>Explanation:</h2>

The elastic potential energy of an elastic material (e.g a spring, a wire), is the energy stored when the material is stretched or compressed. It is given by

U = $\frac{1}{2}kx^2$ --------------------(i)