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USPshnik [31]
3 years ago
14

A _____ has to do with the direction of a force​

Physics
2 answers:
eimsori [14]3 years ago
6 0
In science, work is defined as the action that results when a force causes an object to move in the direction of the force. ... The object moves as a force is applied. The direction of the object's motion is the same as the direction of the force applied.
Elenna [48]3 years ago
3 0

Answer:

The "solid force"? ... The direction of the force always seems to be coming out of the solid surface. A direction which is perpendicular to the plane of a surface is said to be normal. The force that a solid surface exerts on anything in the normal direction is called the normal force.

Explanation:

i think i hope this helps

You might be interested in
Two boxers are fighting. Boxer 1 throws his 5 kg fist at boxer 2 with a speed of 9 m/s.
Sladkaya [172]

Answer:

0.001 s

Explanation:

The force applied on an object is equal to the rate of change of momentum of the object:

F=\frac{\Delta p}{\Delta t}

where

F is the force applied

\Delta p is the change in momentum

\Delta t is the time interval

The change in momentum can be written as

\Delta p=m(v-u)

where

m is the mass

v is the final velocity

u is the initial velocity

So the original equation can be written as

F=\frac{m(v-u)}{\Delta t}

In this problem:

m = 5 kg is the mass of the fist

u = 9 m/s is the initial velocity

v = 0 is the final velocity

F = -45,000 N is the force applied (negative because its direction is opposite to the motion)

Therefore, we can re-arrange the equation to solve for the time:

\Delta t=\frac{m(v-u)}{F}=\frac{(5)(0-9)}{-45,000}=0.001 s

4 0
3 years ago
Water flows through a garden hose which is attached to a nozzle. The water flows through hose with a speed of 1.81 m/s and throu
Serggg [28]

Answer:

a) 17.086m

b) 0.1671 m

Explanation:

Given data: speed of water through the hose  = 1.81 m/s

through the nozzle = 18.3 m/s

We know that maximum height of an object with upward velocity v is given by,

a) H = v^2/2g

where H is the maximum height water emerges  

= 18.3^2/(2×9.8) = 17.086 m answer

b) Again,  

H = v^2/2g

= 1.81^2/(2×9.8) = 0.1671 m

6 0
3 years ago
Computes average speed when a person walk for 1 minute at speed of 1.5 m/s and another 1 minute at speed of 3.5 m/s along a stra
Vinvika [58]

Answer:

2.5 m/s

Explanation:

(60*1.5+60*3.5)/120=2.5

or

(1.5+3.5)/2=2.5

3 0
3 years ago
We can study how galaxies evolve because ______. galaxies are transparent to visible light. we can watch as they interact in rea
Verizon [17]
Here is the answer that completes the statement above. We can study how galaxies evolve because THE FARTHER AWAY WE LOOK, THE FURTHER BACK IN TIME WE SEE. This means that the more we discover more about what's happening in the universe, the more we become curious to know how and when it began. Hope this helps.
6 0
3 years ago
A 0.453 kg pendulum bob passes through the lowest part of its path at a speed of 2.58 m/s. What is the tension in the pendulum c
harkovskaia [24]

Answer with Explanation:

Mass of pendulum bob, m=0.453 kg

Speed, v_1=2.58 m/s

a.r=75.1 cm=75.1\times 10^{-2}m=0.751 m

1cm=10^{-2} m

Tension in the pendulum cable is given  by

Tension=Centripetal force+force due to gravity

T=\frac{mv^2}{r}+mg

Where g=9.8 m/s^2

Substitute the values

T=\frac{0.453(2.58)^2}{75.1\times 10^{-2}}+0.453\times 9.8

T=8.45 N

b.When the pendulum reaches its highest point,then

Final velocity, v_2=0

According to law of conservation of energy

mgh_1+\frac{1}{2}mv^2_1=mgh_2+\frac{1}{2}mv^2_2

gh_1+\frac{1}{2}v^2_1=gh_2+\frac{1}{2}v^2_2

h_1=0

Substitute the values

9.8\times 0+\frac{1}{2}(2.58)^2=9.8\times h_2+\frac{1}{2}(0)^2

3.3282=9.8h_2

h_2=\frac{3.3282}{9.8}=0.34 m

The angle mad  by cable with the vertical=cos\theta=\frac{0.751-0.34}{0.751}=0.55

\theta=cos^{-1}(0.55)=56.6^{\circ}

c.When the pendulum reaches at highest point then

Acceleration, a=0

Therefore, the tension  in the pendulum cable

T=mgcos\theta

Substitute the values

T=0.453\times 9.8cos56.6

T=2.4 N

8 0
4 years ago
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