A(n) 1946 kg car travels at a speed of 10 m/s . What is its kinetic energy ? Answer in units of J.

What is difference between non uniform and uniform circular motion?

statuscvo [17]

Answer:

The object in a uniform motion covers same distances in an equal time period. Objects in a non-uniform motion cover dissimilar distances in an equal time period.

Explanation:

The speed of the object traveling in uniform motion is constant, the actual speed and the average speed of the moving body is same.

6 0

2 years ago

Assuming a typical efficiency for energy use by the body, how many slices of pizza must you eat to walk for 2.5 h at a speed of

larisa86 [58]

Answer:

2.7 Pizzas.

Explanation:

The power required to walk through 5km in 1 hour is 380W.

A watt is basically Jules per second, then we need to standardized this measurement to second.

5km/hr is equal to,

$\frac{5km}{hr}*\frac{1hr}{3600s}*\frac{1000m}{1km}=1.389m/s$

Walking by 2.5 hours is equal to a distance of,

$d=v*t=1.389*(2.5*3600) = 12500m$

The total energy required then would be,

$E = \frac{380J}{1.389m/s}(12500)=3.4199*10^6J$

Then we know that one pizza slice gives $1260*10^3J$ of energy, the total pizza needed are,

$\eta = \frac{3.4199*10^6}{1260*10^3} = 2.7142$

<em>Then you need to buy 3 pizza.</em>

6 0

3 years ago

Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

pantera1 [17]

Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$

The sum of the vertical components of F1 and F2 must equal the total force Ft

$\displaystyle -2F\ cos21^o+F\ sin\alpha =460$

Solving for $\alpha$ in the first equation

$\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o$

$\displaystyle cos\alpha =0.717=>\alpha =44.214^o$

$\displaystyle F(2cos21^o+sin\alpha)=460$

$\displaystyle F=\frac{460}{2cos21^o+sin\alpha}$

$\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}$

$\displaystyle F=179.37\ N$

The magnitude of F1 is

$|F1|=2*F=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

4 0

3 years ago

Protons and neutrons in an atom are held together by _____.

marysya [2.9K]

Protons and neutrons in an atom are held together by a nuclear energy also called the strong force.

7 0

3 years ago

A 0.10 g honeybee acquires a charge of +23 pC while flying.

kari74 [83]

Answer:

a) $\frac{F}{w} =2.347\times 10^{-6}\ N$

b) $E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, $m=10^{-4}\ kg$

charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

a.

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

$F=E.q_2$

$F=100\times 23\times 10^{-12}$

$F=23\times 10^{-10}\ N$

The weight of the bee:

$w=m.g$

$w=10^{-4}\times 9.8$

$w=9.8\times10^{-4}\ N$

Therefore the ratio :

$\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}$

$\frac{F}{w} =2.347\times 10^{-6}\ N$

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

$E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

3 0

2 years ago