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3 years ago

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A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H

ELP

1 answer:

AlexFokin [52]3 years ago

7 0

Answer:

4982.8 W

Explanation:

Given the mass of the car = 1430 kg

initial speed = 7.5m/sec

final speed = 11.0 m/sec

time = 9.30sec

Assuming constant accelration a We know that

v=u+at

11=7.5+9.3a

a=0.376 $m/sec^2$

We know that F=ma

F= $1430*0.376 = 538.2N$

We know that $v^2-u^2=2as$

$11^2-7.5^2=2*0.376*s$

s= 86.10m

We know that work(W)= $F*s$

W= $538.2*86.10$ = 46339.02J = 46.34KJ

We know that power= $\frac{work}{time}$ = $\frac{46.34}{9.3}=4.982$ KW

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Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra

Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

$s^2 = (x_A - x_B)^2+(y_A - y_B)^2$ (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

$s^2 = x_B^2+y_A^2$ (2)

Taking the differential with respect to time:

$\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}$ (3)

where $\displaystyle{\frac{dx_B}{dt}}=v_B$ and $\displaystyle{\frac{dx_A}{dt}}=v_A$ are the respective given velocities of the boats. To find $s$ and $x_B$ we make use of the given position for A, $y_A=30$ , the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

$\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h$

with this time, we know can now calculate the distance at which B is:

$\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi$

and applying Pythagoras:

$\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}$

Now substituting all the values in (3) and solving for $\displaystyle{\frac{ds}{dt} }$ we get:

$\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}$

4 0

3 years ago

WILL give brainliest!!!!

Lostsunrise [7]

People who have been struck by lightning do not carry an electrical charge and can be touched is True.

heat lightning is lightning flashes in the sky that do not have the accompanying sound of thunder so false just like the others

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3 years ago

An 80.0 kg hiker walks a distance of 400.0 m along a road that slopes 5.0 degrees upward, and then stops. What is the hiker's fi

lana66690 [7]

The height difference is found by
$\delta H=400sin(5 \°)=34.86m$
Then the change in potential energy is
$E=mgh=(80.0kg)(9.8 \frac{kgm}{s^2})(34.86)= 27332J$

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3 years ago

A cue ball, moving with 9.0 N·s of momentum strikes the nine-ball at rest. The nine-ball moves off with 2.0 N·s in the original

lisov135 [29]

Answer:

P = 7.28 N.s

Explanation:

given,

initial momentum of cue ball in x- direction,P₁ = 9 N.s

momentum of nine ball in x- direction, P₂ = 2 N.s

momentum in perpendicular direction i.e. y - direction,P'₂ = 2 N.s

momentum of the cue after collision = ?

using conservation of momentum

in x- direction

P₁ + p = x + P₂

p is the initial momentum of the nine balls which is equal to zero.

9 + 0 = x + 2

x = 7 N.s

momentum in x-direction.

equating along y-direction

P'₁ + p = y + P'₂

0 + 0 = y + 2

y = -2 N.s

the momentum of the cue ball after collision is equal to resultant of the momentum .

$P = \sqrt{x^2+y^2}$

$P = \sqrt{7^2+(-2)^2}$

P = 7.28 N.s

the momentum of the cue ball after collision is equal to P = 7.28 N.s

7 0

2 years ago

if a Firebird travels at a velocity of 0 to 60 mph in four seconds traveling east what was the acceleration of the Firebird

Tresset [83]

Answer:

6.7 m/s^2

Explanation:

The formula of acceleration is:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{v_2 - v_1}{t_2-t_1}}$

where $\displaystyle{\vec{a}}$ is acceleration, $\displaystyle{\vec{v}}$ is velocity and $\displaystyle{t}$ is time. $\displaystyle{v_2}$ means final velocity. $\displaystyle{v_1}$ means initial velocity, $\displaystyle{t_2}$ means final time and $\displaystyle{t_1}$ means initial time.

We are given that the Firebird travels at velocity of 0 to 60 mph in four seconds. Therefore:

Our initial velocity starts at 0 mph.
Our final velocity is at 60 mph.
Our initial time is 0 second.
Our final time is 4 seconds.

Since it travels to the east then our vector will be positive. However, acceleration has to be in m/s^2 unit (Sl unit) so we'll have to convert from mph (miles per hours) to m/s (meters per second) first.

We know that:

A mile equals to 1609.344 meters.
An hour equals to 60 minutes which a minute equals to 60 seconds. So 60 minutes will equal to 3600 seconds.

Now we divide 1609.344 by 3600 to find a unit rate of m/s:

$\displaystyle{\dfrac{1609.344}{3600} \ \, \sf{m/s}}\\\\\displaystyle{= 0.44704 \ \, \sf{m/s}}$

Now multiply 0.44704 m/s by 0 and 60 to get velocity in m/s unit:

Initial velocity = 0 m/s
Final velocity = 60 * 0.44704 = 26.82 m/s

Time is already in second so no need for conversion. Substitute known information in the formula:

$\displaystyle{\vec{a} = \dfrac{26.82-0}{4-0}}\\\\\displaystyle{\vec{a} = \dfrac{26.82}{4}}\\\\\displaystyle{\vec{a} = 6.7 \ \, \sf{m/s^2}}$

Therefore, the Firebird will accelerate at the rate of 6.7 m/s^2.

3 0

1 year ago