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Natasha_Volkova [10]

3 years ago

12

A 1800-kg Jeep travels along a straight 500-m portion of highway (from A to B) at a constant speed of 10 m/s. At B, the Jeep enc

ounters an unbanked curve of radius 50 m. The Jeep follows the road from B to C traveling at a constant speed of 10 m/s while the direction of the Jeep changes from east to south. What is the magnitude of the frictional force between the tires and the road as the Jeep negotiates the curve from B to C?

1 answer:

vladimir1956 [14]3 years ago

4 0

Answer:

3600N

Explanation:

$F=m\frac{v^2}{r}$

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Difference between diurnal and annual motion in two points

Valentin [98]

Answer:

Explanation below:

Explanation:

Annual motion describes the changes in motion of the earth around the sun. Diurnal motion can be better understood as the change in motion caused by Earths rotation at the poles.

This might not be the answer you were looking for, your question is very vague.

5 0

3 years ago

An object is moving with a constant velocity of 278 m/s. How long will it take it to travel 7500 m, using the formula Delta X=Vt

rusak2 [61]

If the velocity is constant then the acceleration of the object is zero.
$a=0 (m/s^2)$
Thus when we apply the equation
$\Delta X=vt+(at^2/2)$
It remains
$\Delta X =vt$
or equivalent
$t=(\Delta X/v) =7500/278 =26.98 (seconds)$

7 0

4 years ago

Suppose that you are standing on a train accelerating at 0.20g (where g is the acceleration due to gravity). What minimum coeffi

prisoha [69]

Answer:

0.2

Explanation:

The given parameters are;

The acceleration of the train, a = 0.2·g

The mass of the person standing on the train = m

Let μ represent the coefficient of static friction, we have;

The force acting on the person, F = m × a = m × 0.2·g

The force of friction acting between the feet and the floor, $F_f$ = m·g·μ

For the person not to slide we have;

The force acting on the person = The force of friction acting between the feet and the floor

F = $F_f$

∴ m × 0.2·g = m·g·μ

From which we get;

0.2 = μ

The coefficient of static friction that must exist between the feet and the floor if the person is not to slide, μ = 0.2.

7 0

3 years ago

An intravenous (IV) system is supplying saline solution to a patient at the rate of 0.06 cm3/s through a needle of radius 0.2 mm

horsena [70]

Answer:

Pressure applied to the needle is 7528 Pa

Explanation:

As we know by poiseuille's law of flow of liquid through a cylindrical pipe

the rate of flow through the pipe is given as

$Q = \frac{\Delta P \pi r^4}{8\eta L}$

now we know that

$Q = 0.06 \times 10^{-6} m^3/s$

radius = 0.2 mm

Length = 6.32 cm

$\eta = 1\times 10^{-3} Pa s$

now we have

$6 \times 10^{-8} = \frac{\Delta P \pi (0.2 \times 10^{-3})^4}{8(1 \times 10^{-3})6.32 \times 10^{-2}}$

$3.03 \times 10^{-11} = \Delta P 5.02 \times 10^{-15}$

$\Delta P = 6028 Pa$

now we have

$P - 1500 = 6028 Pa$

$P = 7528 Pa$

8 0

3 years ago

A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves

expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

$q_1=3.4\mu C=3.4\times 10^{-6} C$

$1\mu C=10^{-6} C$

$q_2=-4.9\mu C=-4.9\times 10^{-6} C$

$x_1=0.125,y_1=0$

$x_2=0.280,y_2=0.235$

We have to find the work done by the electric force on the moving point charge.

$r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125$

$r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366$

Work done, $W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})$

Where $k=9\times 10^9$

Using the formula

$W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})$

$W=-0.79 J$

5 0

4 years ago