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NARA [144]
3 years ago
12

A 250 kg motorcycle starts at rest at a stoplight. When the light turns green the motorcycle accelerates to 20 m/s in 4 seconds.

Calculate the force applied by the motorcycle if there is also 200 N of friction. Please show work. Any help would be appreciated
Physics
1 answer:
aliya0001 [1]3 years ago
6 0
F=MA
F + 200= 250 x 4
F + 200 = 1000
F= 1000-200
F= 800
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3 years ago
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A car is cruising at a steady speed of 35 mph. Suddenly, a cuddly puppy runs out into the road. The driver takes 1.7 seconds to
Schach [20]

Answer:

The distance traveled is 0.037 mi

Explanation:

The equation for the position and velocity of an accelerated object is:

x = x0 + v0 * t + 1/2 * a * t²

v = v0 + a * t

where

x = position at time t

x0 = initial position

t = time

a = acceleration

v0 = initial velocity

If the velocity is constant, then a = 0 and the position will be:

x = x0 + v * t where "v" is the velocity

First, let´s find the distance traveled until the driver push the brake:

The speed is constant. Then:

x = x0 + v * t (considering the origin of the reference system to be located at the point at which the driver sees the puppy, x0 = 0)

x = 35 mi/h (1 h / 3600 s) * 1.7 s = 0.017 mi

Then, the drivers moves with constant acceleration until the car stops (v = 0)

From the equation for velocity:

v = v0 + a * t

Since v = 0, we can obtain the acceleration of the car until it stops. With that acceleration, we can calculate how much distance the car moves before it stops.

0 = v0 + a * t

-v0 / t = a

-35 mi/h (1 h / 3600s) / 4.0 s = a

a = -2.4 x 10⁻³ mi/s²

The distance traveled will be:

x = x0 + v0 * t + 1/2 * a * t²

Now x0 will be the distance traveled before the driver slows down.

x = 0.017 mi + 35 mi/h (1 h / 3600s) * 4 s + 1/2 * ( -2.4 x 10⁻³ mi/s²) * (4s)²

x = 0.037 mi

6 0
3 years ago
A wooden block contains some nails so that its density is exactly equal to that of water. If it is placed in a tank of water and
belka [17]

Answer:

The correct answer is:

a) remain where it is released

Explanation:

The concept of density seeks to measure the weight of an object in relation to its size. It is the measure of how packed together the particles of that object are. An object placed in a liquid displaces a certain volume of the liquid, based on the relative density of the object and the liquid.

If an object is less dense than a liquid in which it is placed, it displaces a smaller volume of the liquid than its volume, hence only some part of the object will be seen to be under the liquid, the other part will float.

If an object is denser than the liquid in which it is placed, it displaces a larger volume of the liquid than its own volume, making the object to sink and is submerged, sometimes to the bottom of the liquid, but mostly below the point at which it was released.

Finally, if the density of an object and the liquid into which it is submerged is the same. the object's mass per unit volume is the same as the liquid's mass per unit volume, hence the weight and force created due to density will balance and cancel each other out hence making the object to remain where it was submerged.

8 0
3 years ago
calcula la potencia por hora de un radiador, sabiendo que esta conectado a un contacto común 110 v. y requiere 20 Amp.
Ira Lisetskai [31]

calculate the power per hour of a radiator, knowing that it is connected to a common 110 v contact. and requires 20 Amp.

Answer:

2.2kWh

Explanation:

Given parameters:

Potential difference  = 110v

Current  = 20A

Unknown:

Power  = ?

Solution:

To solve this problem, we use the expression below:

        Power  = IV

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  This is therefore 2.2kW

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8 0
3 years ago
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oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

Q = FA\sigma\Delta T^4

Where,

F =View Factor

A = Cross sectional Area

\sigma = Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K

The view factor between two coaxial parallel disks would be

\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33

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Then the view factor between base to top surface of the cylinder becomes F_{12} = 0.26. From the summation rule

F_{13} = 1-0.26

F_{13} = 0.74

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}

\dot{Q_3} = 2\dot{Q_{13}}

\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)

\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)

\dot{Q_3} = 780.76W

Therefore the rate heat radiation is 780.76W

5 0
3 years ago
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