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2 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]2 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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__________ are scientists that study the composition of earth including rock layers at the surface.

insens350 [35]

Answer:

Geology is the study of planet Earth, including its composition and structure. Geologists are scientists who study Earth and the processes that shape Earth over time. Geologists study two types of forces that change Earth's surface.

Explanation:

please mark me brainliest

4 0

2 years ago

Can someone help me with this Physics question please?

strojnjashka [21]

Answer:

Explanation:

The formula for this, the easy one, is

$N=N_0(\frac{1}{2})^{\frac{t}{H}$ where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:

$N=48.0(\frac{1}{2})^{\frac{49.2}{12.3}$ and simplifying a bit:

$N=48.0(.5)^4$ and

N = 48.0(.0625) so

N = 3 mg left after 12.3 years

6 0

2 years ago

Read 2 more answers

You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body

stellarik [79]

A) The change in internal chemical energy is $1.15\cdot 10^7 J$

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

$P=Fv$

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

$P=(80)(8.0)=640 W$

The energy output is related to the power by the equation

$P=\frac{E}{t}$

where:

P = 640 W is the power output

E is the energy output

$t = 30 min \cdot 60 = 1800 s$ is the time elapsed

Solving for E,

$E=Pt=(640)(1800)=1.15\cdot 10^6 J$

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

$\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J$

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

$E=1.15\cdot 10^7 J$

Here we want to find the time t' needed to convert an amount of chemical energy of

$E'=3.8\cdot 10^5 J$

So we can setup the following proportion:

$\frac{t}{E}=\frac{t'}{E'}$

And solving for t',

$t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min$

Learn more about power and energy:

brainly.com/question/7956557

#LearnwithBrainly

3 0

2 years ago

1. Bone has a Young’s modulus of about

Blababa [14]

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

$strain = 9.33 \times 10^{-3}$

now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

$\rho = 1185.3 kg/m^3$

as we know that pressure changes with depth as per following equation

$P = P_o + \rho g h$

here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

$2P_0 = \rho g h$

$2(1.01 \times 10^5) = 1025 (9.81)(h)$

here we will have

$h = 20.1 m$

so it is 20.1 m below the surface

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

$mg = \rho_1V_1g + \rho_2V_2g$

$A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)$

$46.6 = 43.89 - 922x + 1000x$

$x = 3.48 cm$

so it is 3.48 cm below the interface

5 0

3 years ago

A spring with spring constant of 34 N/m is stretched 0.12 m from its equilibrium position. How much work must be done to stretch

Nesterboy [21]

Answer:0.253Joules

Explanation:

First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.

F = ke where;

F is the force

k is spring constant = 34N/m

e is the extension = 0.12m

F = 34× 0.12 = 4.08N

To get work done,

Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.

Work done = Force × Distance

Since F = 4.08m, distance = 0.062m

Work done = 4.08 × 0.062

Work done = 0.253Joules

Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules

8 0

3 years ago