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padilas [110]
2 years ago
8

an 2-kg object is moving horizontally with a speed of 4m/s. how much net force os required to keep the object movong with the sa

me speed and in the same direction

Physics
1 answer:
Zepler [3.9K]2 years ago
7 0
In order to keep the object moving in the same direction and with same speed, its initial and final momentum should be zero.
∴ Net force on the object should be zero.
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A person travels distance πR along the circumference
lakkis [162]

Answer: 2R

Explanation:

Here the person travels пR distance. We know that the circumference of a circle is 2πR. So your imaginated person has traveled the distance which​ is half of the circumference of the circle. And this distance is equal to its diameter. We know that diameter of a circle is two times larger than the radius. So the person's displacement is two times of the radius, means 2R. [Here 'R' means the radius of the circle]

8 0
2 years ago
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An green hoop with mass mh = 2.8 kg and radius rh = 0.13 m hangs from a string that goes over a blue solid disk pulley with mass
Otrada [13]
The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table. 
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. 
I = 7/5*mR^2 M = 7/5*m 
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2 
7 0
2 years ago
What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22
SIZIF [17.4K]

Answer:

5.45\times 10^{-4} W

Explanation:

T_{r} = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

T_{s} = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

A = Surface area = 1.50 m²

\epsilon = emissivity = 0.97

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴

Rate of heat transfer is given as

R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})

R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})

R = 5.45\times 10^{-4} W

3 0
2 years ago
A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c
Rzqust [24]

Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

  • The plane wall is initially with constant properties is initially at a uniform temperature, To.
  • Suddenly the surface x=L is exposed to convection process such that T∞>To.
  • The other surface x=0 is maintained at To
  • Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

  1. There is only conduction in 1-D.
  2. The system bears constant properties.
  3. The volumetric heat generation is uniform

From the given data, the condition are as follows

<u>Initial Condition</u>

At t≤0

T(x,0)=T_o

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

<u>Boundary Conditions</u>

<u>At x=0</u>

<u />T(0,t)=T_o<u />

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

<u>At x=L</u>

<u />-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]<u />

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

3 0
2 years ago
A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.
Makovka662 [10]

Answer:

a)a=5 i+2t j - 6\ t^2k

b)a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

Explanation:

Given that

v(t) = 5 t i + t² j - 2 t³ k

We know that acceleration a is given as

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k

a=5 i+2t j - 6\ t^2k

Therefore the acceleration function a will be

a=5 i+2t j - 6\ t^2k

The acceleration at t = 2 s

a= 5 i + 2 x 2 j - 6 x 2² k  m/s²

a=5 i + 4 j -24 k m/s²

The magnitude of the acceleration will be

a=\sqrt{5^2+4^2+24^2}\ m/s^2

a= 24.83 m/s²

The direction of the acceleration a is given as

a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

a)a=5 i+2t j - 6\ t^2k

b)a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

5 0
2 years ago
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