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marshall27 [118]
3 years ago
6

A sled of mass 10 kg slides along the ice. it has an initial speed of 2 m/s but stops because of friction. How much work is done

by friction?
A. 0 J
B. -20 J
C. -10 J
D. 20 J
Physics
2 answers:
NeX [460]3 years ago
6 0

Answer: The correct answer is option B.

Explanation:

Mass of the sled = 10 kg

Initial speed of the sled = 2 m/s

Kinetic energy of the sled = \frac{1}{2}mv^2

\frac{1}{2}\times 10 kg\times (2 m/s)^2=20 Joules

Work done by the sled = 20 joules

The work done by the friction will be in opposite direction and equal to the magnitude of the work done of the sled that - 20 J.

Hence, correct answer is option B.

Alexandra [31]3 years ago
5 0

Answer:-20

Explanation:

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How long will it take for a boat traveling at 36 k/h to go 126 km?
brilliants [131]
3 hours 30 min i believe because 126/36 is 3.5 and 3.5 hours is 3 hours and 30 min. Brainliest pls
5 0
3 years ago
Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

5 0
3 years ago
2)
earnstyle [38]

Answer:

0.0312J

Explanation:

Let x be the distance the staple moves:

x=0.150m-0.115m=0.035m

And spring constant is k=51.0N/m

PE=0.5kx^2\\=0.5\times 51.0\times 0.035^2\\\\=0.312

Hence, the potential energy is 0.0312J

8 0
3 years ago
3. What color of laser light shines through a diffraction grating with a line density of 500 lines/mm if the third maxima from t
mestny [16]

Answer:

Wavelength is 471 nm

Explanation:

Given that,

Lines per unit length of diffraction grating is 500 lines/mm.

The third maxima from the central maxima (m=3) is at an angle of 45°

We need to find the color of laser light shines through a diffraction grating.

The condition for maxima is :

d\sin\theta=m\lambda

d = 1/N, N = number of lines per mm

\lambda=\dfrac{1}{Nm}\sin\theta\\\\\lambda=\dfrac{10^{-3}}{500\times 3}\sin(45)\\\\\lambda=4.31\times 10^{-7}\\\\\text{or}\\\\\lambda=471\ nm

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A spring of mass
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