Anyones good in gr5 science pls help me

During a field experiment about speed, a scientist created the chart above. The chart shows distance and time measurements for a

BaLLatris [955]

Answer:

yes

Explanation:

its not a good thing for the rest of your life but you have a PS4

6 0

3 years ago

An archer shoots an arrow with a mass of 45.0 grams from bow pulled

Sladkaya [172]

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

3 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

an 1150kg elevator moving down speeds up at a rate of 3.5m/s. what is the tension in the supporting cables?

gtnhenbr [62]

Answer:

The tension force in the supporting cables is 7245N

Explanation:

There are two forces acting on the elevator: the force of gravity pointing down (+) with magnitude (elevator mass) x (gravitational acceleration), and the tension force of the cable pointing up (-) with an unknown magnitude F. The net force is the sum of these forces:

$F_{net} = F_g - F = m\cdot g - F\\$