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otez555 [7]
2 years ago
15

On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max

pulls on the sled with 12 n of force, directed at an angle of 15° above the ground. when he comes to a recently plowed section of road, he continues to pull the sled with the same force across the road while the road exerts a frictional force of 4 n on sled. what is the net work done on the sled while max pulls it 5 m across the road?

Physics
1 answer:
sesenic [268]2 years ago
7 0

Work done by a given force is given by

W = F.d

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = W_1 = Fdcos\theta

W_1 = 12*5cos15

W_1 = 57.96 J

Now similarly work done by frictional force

W_2 = Fdcos\theta

W_2 = 4*5cos180

W_2= -20 J

Now total work done on sled

W_{net}= W_1 + W2

W_{net} = 57.96 - 20 = 37.96 J

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The blood plays an important role in removing heat from th ebody by bringing the heat directly to the surface where it can radia
BabaBlast [244]

Answer: Thermal comductivity (K) is 3.964x 10 ^-3 W/m.k

Explanation:

Thermal comductivity K = QL/A∆T

Q= Amount of heat transferred through the material in watts = 75W

L= Distance between two isothermal planes = 0.740mm

A= Area of the surface in square metres = 2m^2

∆T= Temperature change = (37-30) °C.

Solving this : K =( 75 x 0.740 x 10^-3)/ 2 x (37-30)

K = 3.964x 10 ^-3 W/m.k

7 0
2 years ago
Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
kodGreya [7K]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

7 0
2 years ago
The headlights of a car emit light of wavelength 400 nm and are separated by 1.2 m. The headlights are viewed by an observer who
densk [106]

Answer:

The most correct option is;

B. 10 km

Explanation:

L = \frac{y \times d}{1.22 \times  \lambda} = \frac{1.2 \times 0.004}{1.22 \times  400 \times 10^{-9}} = 9836.066 \ km

Where:

y = Distance between the two headlights

d = Aperture of observers eye

λ = Wavelength of light

L = Distance between the observer and the headlight

Therefore, from the above solution, the distance between the observer and the headlights is 9386.066 km which is approximately 10 km.

Also we have

sinθ = y/L = 1.22 (λ/d)  

= 1.22 \times \frac{400 \times 10^{-9}}{0.004}

sinθ = 1.22×10⁻⁴ rad

6 0
2 years ago
Read 2 more answers
Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu
Veronika [31]

Answer:

Jari

Explanation:

The question requires to know who is traveling faster. This is done by comparing the gradients. The steeper the slope (high gradient), the faster the speed and vice versa.

From Jari's line, the starting point is (0, 0) and another point is (6, 7)

The gradient being change in y to change in x

Change in y=7-0=7

Change in x=6-0=6

Slope is 7/6

For Jade, first point is (0, 10) then another point is (6, 16)

Change in y=16-10=6

Change in x=6-0=6

Slope is 6/6=1

Clearly, 7/6 is greater than 6/6 or 1 hence Jari is faster than Jade

3 0
2 years ago
A 30-km, 34.5-kV, 60-Hz, three-phase line has a positive-sequence series impedance z 5 0.19 1 j0.34 V/km. The load at the receiv
zmey [24]

Answer:

(a) With a short line, the A,B,C,D parameters are:

    A = 1pu    B = 1.685∠60.8°Ω    C = 0 S    D = 1 pu

(b) The sending-end voltage for 0.9 lagging power factor is 35.96 KV_{LL}

(c) The sending-end voltage for 0.9 leading power factor is 33.40 KV_{LL}

Explanation:

(a)

Considering the short transition line diagram.

Apply kirchoff's voltage law to the short transmission line.

Write the equation showing the relations between the sending end and the receiving end quantities.

Compare the line equations with the A,B,C,D parameter equations.

(b)

Determine the receiving-end current for 0.9 lagging power factor.

Determine the line-to-neutral receiving end voltage.

Determine the sending end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

(c)

Determine the receiving-end current for 0.9 leading power factor.

Determine the sending-end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

8 0
2 years ago
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