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2 years ago

8

a motorcycle begins at rest and accelerates uniformly at 7.9 m/s2. We want to find the time it takes the motorcycle to reach a s

peed of 100 km/h. Which kinetic formula would be most useful to solve for the target unknown?

2 answers:

gladu [14]2 years ago

6 0

Answer:

It takes the motorcycle about 12.65 seconds to reach 100mph

Explanation:

AURORKA [14]2 years ago

3 0

Answer:3.5 seconds

Explanation:

initial velocity=u=0

Final velocity=v=100km/h

Final velocity=v=(100x5)/18=500/18=27.8 m/s

Acceleration=a=7.9m/s^2

t=(v-u)/a

t=(27.8-0)/7.9

t=27.8/7.9

t=3.5 seconds

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In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into t

Pie

Answer:

A) $Q_a=74256\ J$

B) $Q=93562560\ J$

Explanation:

Given:

temperature of air, $T_a=-19+273=254\ K$

temperature of lungs, $T_l=37+273=310\ K$

specific Heat exchanged from the lungs , $c_l=0.47\ J.kg^{-1}.K^{-1}$

specific heat of air, $c_a=1020\ J.kg^{-1}.K^{-1}$

mass of 1 L air, $m'=1.3\ kg$

breath rate, $b=21\ breath.min^{-1}$

A)

Now,

amount of heat needed to warm the air of lungs to the body temperature:

$Q_a=m'.c_a.\Delta T$

$Q_a=1.3\times1020\times (310-254)$

$Q_a=74256\ J$

B)

Amount of heat lost per hour:

<u>No. of breaths per hour:</u>

$B=b.60$

$B=21\times 60$

$B=1260$

<u>Now the total loss of energy in 1 hr.:</u>

$Q=Q_a.B$

$Q=74256\times 1260$

$Q=93562560\ J$

7 0

2 years ago

A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t

Pie

Answer:

$I = \frac{mvb}{6}$

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

$L = I$ ω .... (1)

moment of inertia of rod rotating about its center of length b

$I = \frac{ mb^2}{12}$ ........ .(2)

using v = ωr

where w is angular velocity

and r is radius of rod which is equal to b

so we get 2v = ωb

ω = 2v/b ................. (3)

here velocity is two time because two opposite ends are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

$L = \frac{mb^2}{12}$ × $\frac{2v}{b}$

so final answer will be $L = \frac{mvb}{6}$

7 0

2 years ago

Observe: Up until now, all the problems you have solved have involved converting only one unit. However, some conversion problem

ipn [44]

Answer:

t is appropriate to clarify that units such as time and angles the transformation is not in base ten, for example:

60 s = 1 min

60 min = 1 h

24 h = 1 day

Therefore, for this transformation, you must be more careful

the length transformation is base 10

Explanation:

In many exercises the units used are transformed by equations into other units called derivatives, in general the transformation of derived units is the product of the transformation of the constituent units.

In the example of velocity, the derivative unit is m / s, which is why it works in the same way that you transform length and time if in the equation it is multiplying it is multiplied and if it is dividing it is divided.

It is appropriate to clarify that units such as time and angles the transformation is not in base ten, for example:

60 s = 1 min

60 min = 1 h

24 h = 1 day

Therefore, for this transformation, you must be more careful

the length transformation is base 10

1000 m = 1 km

7 0

2 years ago

A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and

snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

$\dfrac{dp}{dt}=F$

$\dfrac{dP}{dt}=\dfrac{mv^2}{r}$

Put the value into the formula

$\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}$

$\dfrac{dP}{dt}=794.11\ N$

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0

2 years ago

A bird is flying in a room with a velocity field of . Calculate the temperature change that the bird feels after 9 seconds of fl

Korvikt [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is $\frac{dT}{dt} = 1.016 ^oC/m$

Explanation:

From the question we are told that

The velocity field with which the bird is flying is $\vec V = (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s$

The temperature of the room is $T(x, y, u) = 400 -0.4y -0.6z-0.2(5 - x)^2 \ ^o C$

The time considered is t = 10 \ seconds

The distance that the bird flew is x = 1 m

Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird

Generally the change in the bird temperature with time is mathematically represented as

$\frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 * (5-x)] [-\frac{dx}{dt} ]$

Here the negative sign in $\frac{dx}{dt}$ is because of the negative sign that is attached to x in the equation

So

$\frac{dT}{dt} = -0.4v_y -0.6v_z -0.2[2 * (5-x)][ -v_x]$

From the given equation of velocity field

$v_x = 0.6x$

$v_y = 0.2t$

$v_z = -1.4$

So

$\frac{dT}{dt} = -0.4[0.2t] -0.6[-1.4] -0.2[2 * (5-x)][ -[0.6x]]$

substituting the given values of x and t

$\frac{dT}{dt} = -0.4[0.2(10)] -0.6[-1.4] -0.2[2 * (5-1)][ -[0.61]]$

$\frac{dT}{dt} = -0.8 +0.84 + 0.976$

$\frac{dT}{dt} = 1.016 ^oC/m$

5 0

2 years ago