3 years ago

Please help me answer this question I will be given the brainliest as well

1 answer:

3 0

draw a horizontal straight line from 25 to 75 and connects the line drawn at 75 to the bottom of the graph (0) at 90 with a straight line

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How much work are you doing if you push on a 40 N rock that won't move?

tangare [24]

If the object doesn't move you have done no work.

5 0

3 years ago

Convert : <br>a) 110°C into °F <br>b) 32°F into °C <br>Please show your work if possible .<br>

Oliga [24]

Answer:

110°C = 230°F

32°F = 0°C

Explanation:

(110*9/5)+32=230

(32-32)*5/9=0

7 0

3 years ago

An object with a mass of 7.60 kg is moving to the right and experiences an applied force of 50 N to the right. The friction forc

horrorfan [7]

Answer:

Explanation:

We will use the equation F - f = ma, which is a fancy way of stating Newton's 2nd Law.

F = +50.0,

f = -30.0,

m = 7.60 kg. Therefore:

50.0 - 30.0 = 7.60a and

20.0 = 7.60a and

a = 20.0/7.60 so

a = 2.63 m/s/s to the right

4 0

3 years ago

A straight trail with a uniform inclination of 15 degrees leads from a lodge at an elevation of 600 feet to a mountain lake at a

9966 [12]

Answer:

The length of the trail = 22796 ft

Explanation:

From the ΔABC

AC = length of the trail = x

AB = 6100 - 600 = 5500 ft

Angle of inclination $\theta$ = 15°

$\sin \theta = \frac{AB}{AC}$

$\sin 15 = \frac{5900}{x}$

$x = \frac{5900}{0.2588}$

x = 22796 ft

Since x = AC = Length of the trail.

Therefore the length of the trail = 22796 ft

7 0

3 years ago

Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8

Kamila [148]

Answer:

34.6 m/s

Explanation:

From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg

Final mass will be 31.5+25.9=57.4 kg

From formula of momentum

M1v1=m2v2

Making v2 the subject of the formula then

$V2=\frac {M1v1}{m2}$

Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then

$V2=\frac {100.2 kg\times 19.8 m/s}{57.4 kg}=34.56376 m/s\approx 34.6 m/s$

4 0

3 years ago