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3 years ago

A student runs 100 meters in 11 seconds. What is the speed of the student?

2 answers:

8 0

Speed has a unit of meters per second in the SI unit system. Speed is the amount of displacement traveled or moved by a moving object or entity per unit time. In this case, we are not limited to only dividing the given data but also converting meters to miles and seconds to hours. The answer to the problem is C . 20.3 miles per hour

GarryVolchara [31]3 years ago

5 0

Speed = distance / time

speed = 100 / 11

speed = 9.09m/s

to get rid of meters and get miles you want to multiply by 1 mile then divide by 1609 meters. to get hours, you times the amount of seconds in an hour (3600) and divide by 1 hour

9.09m/s × 1mile / 1609m × 3600s / 1 hour = 20.3 miles / hour

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It took a bulldozer 62,000 J of work to move a rock 30 m. It took 5 minutes. How much force did the bulldozer have to apply?

NeX [460]

Answer:

A (2066,6 N)

Explanation:

Use the Work formula

62.000J = F . 30

62.000/30 = 2066,6 N

The amout of time it took to move the rock doesn´t matter at all.

It is called a distraction variable, We don´t need it to solve the problem it is there just to confuse.

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3 years ago

.Why does it take double the applied force to move a mass double the size?

Likurg_2 [28]

Answer:

NVIDIA GeForce RTX 3080 10GB GDDR6X PCI Express 4.0 Graphics Card

Explanation:

6 0

2 years ago

If two is company and three is a crowd, what is 4 and 5?

LiRa [457]

4 and 5 is 9

(4 + 5)

4 0

3 years ago

Read 2 more answers

If 1.00 mol of argon is placed in a 0.500-L container at 28.0 ∘C , what is the difference between the ideal pressure (as predict

Rudik [331]

Answer:

1.98 atm

Explanation:

Given that:

Temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28 + 273.15) K = 301.15 K

n = 1

V = 0.500 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

P × 0.500 L = 1 ×0.0821 L atm/ K mol × 301.15 K

⇒P (ideal) = 49.45 atm

Using Van der Waal's equation

$\left(P+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$

R = 0.0821 L atm/ K mol

Where, a and b are constants.

For Ar, given that:

So, a = 1.345 atm L² / mol²

b = 0.03219 L / mol

So,

$\left(P+\frac{1.345\times \:1^2}{0.500^2}\right)\left(0.500-1\times 0.03219\right)=1\times 0.0821\times 301.15$

$P+\frac{1.345}{0.25}=\frac{24.724415}{0.46781}$

$P=\frac{24.724415}{0.46781}-\frac{1.345}{0.25}$

⇒P (real) = 47.47 atm

Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm

4 0

3 years ago

If a box is pushed 5.0 m across a floor using a force of 35 N, how much work was done?

aniked [119]

Answer:

175J

Explanation:

Work done is given by the product of the force applied and the distance travelled as a result of that force.

Or in formula , W=F×d

Thus,

W= 35 × 5.0=175 J or 1.8 e2 J

3 0

3 years ago