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DedPeter [7]
3 years ago
10

A body travels 30m in 5s, 45m in 7s and then 65m in the last 5s. Find the average speed of the body

Physics
1 answer:
JulijaS [17]3 years ago
8 0
We can find the average speed of the body by finding the total distance covered, and then dividing it by the total time of the motion.

The total distance covered is:
S=30 m + 45m+65m=140 m

while the total time of the motion is
t=5 s+7s+5 s=17 s

So, the average speed of the body is:
v= \frac{S}{t}= \frac{140 m}{17 s}=8.24 m/s
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The boiling point for liquid nitrogen at atmospheric pressure is 77k. Is the temperature of an open container of liquid nitrogen
algol13
Answer:
It would be lower than because, if the boiling point of that element is 77 Kelvin degrees then if it isn’t at boiling point it would automatically be cooler than that. Even if it’s at its original state. The normal temperature of Liquid Nitrogen is really cold -320.8 degrees.

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8 0
3 years ago
Lake Baikal in Siberia has a maximum depth of 1642 m. What is the water
Veseljchak [2.6K]

Explanation:

Fluid gauge pressure is:

P = ρgh

where ρ is the fluid density and h is the depth of the fluid.

P = (1000 kg/m³) (9.8 m/s²) (1642 m)

P = 16,091,600 Pa

Rounded to four significant figures, the gauge pressure is 16.09 MPa.

3 0
3 years ago
The distance from Earth to the star Epsilon Eridani is about 10.5 light years. Which of the following statements is true?
Simora [160]
<span>The correct answer is B. - It would take a ray of light 10.5 light years to travel from Earth to Epsilon Eridani, or vice-versa. Using our current technology it would take far longer than 21.0 years for a space ship from Earth to travel that far - I would have to guess many hundreds of years.</span>
4 0
3 years ago
Read 2 more answers
A sculpture is suspended in equilibrium by two cables, one from a wall and the other
aleksandrvk [35]

Answer:

T_1=6655.295917 \approx 6655.3N

Explanation:

From the question we are told that

Angle of cable 2 \theta=37.0\textdegree

Weight of sculpture W=5000 N

Generally the Tension from cable 2 T_2 is mathematically given by

   T_2sin37\textdegree=5000N

   T_2=5000N/sin37\textdegree

   T_2=8308.2N

Generally the Tension from Cable 1 T_1 is mathematically given by

   T_1=T_2 cos37\textdegree

   T_1=8308.2* cos 37\textdegree

   T_1=6655.295917 \approx 6655.3N

7 0
3 years ago
Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0
3 years ago
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