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2 years ago

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A car needs to generate 75.0 hp in order to maintain a constant velocity of 18.2 m/s on a flat road. What is the magnitude of th

e total resistive force acting on the car (due to friction, air resistance, etc.)

1 answer:

stiks02 [169]2 years ago

3 0

Answer:

The magnitude of the force is F= 3072.9 N.

Explanation:

The expression for the power in terms of force and velocity is as follows;

P=Fv

Here, P is the power, F is the force and v is the velocity.

Rearrange the above expression to get the value of the force.

$F=\frac{P}{v}$

Convert power from horsepower to watt.

P=75.0 hp

P= (75)(745.7) W

P= 55,927.5 W

Put P= 55,927.5 W and v= 18.2 meter per second in the above expression.

$F=\frac{55,927.5}{18.2}$

F= 3072.9 N

Therefore, the magnitude of the force is F= 3072.9 N.

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A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o

marshall27 [118]

Answer:

The magnitude of the acceleration of the box is 2 m/s².

Explanation:

Given:

Mass of the box, $m=5.0$ kg

Force acting towards east, $F=27$ N

Frictional force acting towards west, $f=17$ N

Let the acceleration be $a$ m/s².

Now, net force acting on the box towards east is given as:

$F_{net}=F-f=27-17=10\textrm{ N}$

From Newton's second law of motion,

$F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2$

Therefore, the magnitude of the acceleration of the box is 2 m/s².

6 0

2 years ago

Read 2 more answers

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

vredina [299]

Answer:

The angular speed after 6s is $\omega = 1466.67s^{-1}$ .

Explanation:

The equation

$I\alpha = Fd$

relates the moment of inertia $I$ of a rigid body, and its angular acceleration $\alpha$ , with the force applied $F$ at a distance $d$ from the axis of rotation.

In our case, the force applied is $F = 22N$ , at a distance $d = 6cm =0.06m$ , to a ring with the moment of inertia of $I =mr^2$ ; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I}$

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2}$

$\alpha = 244.44\: s^{-2}$

Therefore, the angular speed $\omega$ which is

$\omega = \alpha t$

after 6 seconds is

$\omega = 244.44$\: s^{-2}* 6s$

$\boxed{\omega = 1466.67s^{-1}}$

7 0

2 years ago

A thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center. A force F is exerted tangentially to th

Stella [2.4K]

Given :

Thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center.

A force F is exerted tangentially to the hoop. If the hoop’s radius is 2.0 m and it is rotating with an angular acceleration of 2.5 rad/s².

To Find :

The magnitude of F.

Solution :

Torque on hoop is given by :

$\tau =F\times R\\\\I\alpha = FR\\\\MR^2\alpha = FR\\\\F = MR\alpha$ ( Moment of Inertia of hoop is MR² )

Putting value of M, R and α in above equation, we get :

$F=5\times 2\times 2.5\ N\\\\F = 25 \ N$

Therefore, the magnitude of force F is 25 N.

Hence, this is the required solution.

5 0

1 year ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

2 years ago

The length of a wire 2.00 m is measured as 2.02m. What is the percentage error in the measurement?

n200080 [17]

Answer:

1%

Explanation:

Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.

$Percent Error=\frac{V_{measured}- V_{true} } {V_{true}} *100$

The measured value is 2.02 meters and the actual value is 2.00 meters.

$V_{measured}=2.02\\\\V_{true}=2.00$

$Percent Error=\frac{2.02-2.00}{2.00} *100$

First, evaluate the fraction. Subtract 2.00 from 2.02

$Percent Error=\frac{0.02}{2.00}*100$

Next, divide 0.02 by 2.00

$PercentError=0.01 *100$

Finally, multiply 0.01 and 100.

$Percent Error=1\\Percent Error= 1 \%$

The percent error is 1%.

6 0

2 years ago