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3 years ago

A drive find difficulty to drive a car when the ground is cover with ice explain why?

1 answer:

7 0

The ground is less rough, so there is less friction between the ground and the wheels of the car. This makes it unable for the car to move forward

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A horizontal pipe of inner diameter 2.2 cm carries water with a density of 1000.0 kg/m3 flowing at a rate of 1.5 kg/s. If the pi

EleoNora [17]

The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is, $\rm d_1 = 2.2 \ cm$

initial radius,

$r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm$

The initial crossection area;

$\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2$

The final crossection area;

$\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2$

The initial flow rate is;

R = density ×velocity ×area

$\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1 = 3.947 \ m/sec$

The speed of the water in the wider part will be;

From the continuity equation;

$\rm A_1 V_1 = A_2V_2 \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec$

Hence, the speed of the water in the wider part will be 1.194 m/sec.

To learn more about the speed, refer to the link;

brainly.com/question/7359669

#SPJ1

7 0

1 year ago

Determine the gravitational potential energy, in kJ, of 3 m3 of liquid water at an elevation of 40 m above the surface of Earth.

melomori [17]

Explanation:

We will calculate the gravitational potential energy as follows.

$P.E_{1} = mgz_{1}$

$P.E_{1} = (\rho V)gz_{1}$

= $1000 kg/m^{3} \times 3 m^{3} \times 9.7 \times 40 m$

= 1164000 J

or, = 1164 kJ (as 1 kJ = 1000 J)

Now, we will calculate the change in potential energy as follows.

$\Delta P.E = mg(z_{2} - z_{1})$

= $\rho \times V \times g (z_{2} - z_{1})$

= $1000 \times 3 \times 9.7 (10 - 40)m$

= -873000 J

or, = -873 kJ

Thus, we can conclude that change in gravitational potential energy is -873 kJ.

4 0

2 years ago

If the potential due to a point charge is 490 V at a distance of 10 m, what are the sign and magnitude of the charge?

uysha [10]

Answer:

+5.4×10⁻⁷ C

Explanation:

Electric potential: This can be defined as the work done in bringing a unit charge from infinity to that point against the action of the field. The S.I unit of potential is volt (V)

The formula for potential is

V = kq/r............................ Equation 1

Where V = electric potential, k = proportionality constant, q = charge, r = distance.

making q the subject of the equation,

q = Vr/k............................ Equation 2

Given: V = 490 V, r = 10 m,

Constant: k = 9×10⁹ Nm²/C²

Substitute into equation 2

q = 490(10)/(9×10⁹)

q = 5.4×10⁻⁷ C

q = +5.4×10⁻⁷ C

Hence the charge is +5.4×10⁻⁷ C

7 0

3 years ago

An ice skater is spinning at 6.00 rev/s with his moment of inertia being 0.400 kg/m2. Calculate his new moment of inertia if he

labwork [276]

Answer:

New moment of inertia will be $I=1.92kgm^2$

Explanation:

It is given initially angular velocity $\omega =6rev/sec=6\times 2\pi =37.68rad/sec$

Moment of inertia $I=0.4kgm^2$

Angular momentum is equal to $L=I\omega =37.68\times 0.4=15.072kgm^2/sec$

Now angular velocity is decreases to $\omega =1.25rev/sec=1.25\times 2\times 3.14=7.85rad/sec$

As we know that angular momentum is conserved

So $15.072=I\times 7.85$

$I=1.92kgm^2$

So new moment of inertia will be $I=1.92kgm^2$

4 0

3 years ago

A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude

miskamm [114]

Answer:

Explanation:

Given

Weight of person $W=652\ N$

At highest point Magnitude of the normal force $F=585\ N$

net force at highest point

$F_{net}=W+F_c$

where $F_c=$ centripetal force

$F_c=\dfrac{mv^2}{r}$

$F_{net}=F_{n}=$ Normal Force

$585=652+F_c$

$F_c=-67\ N$

Negative sign shows force is in upward direction

At bottom point centripetal force is towards the bottom

$F_n=F_c+W$

$F_n=652+67$

$F_n=719\ N$

8 0

3 years ago