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weqwewe [10]
3 years ago
8

A 10-kg object is pushed across a rough surface. It begins at rest and accelerates to a speed of 4 m/s in a distance of 5m. Whic

h fact(s) support the conclusion that the object experiences an unbalanced force?
Physics
1 answer:
brilliants [131]3 years ago
5 0

Answer:

Explanation:

Unbalanced Force according to newton's second law is the one which causes  the object to move or break its state of rest on application of force.

Here the object accelerate to a speed of 4 m/s and moves a distance of 5 m on application of force .

Thus we can say that the applied force is unbalanced in nature.

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The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme a
Tanzania [10]

Answer:

1) tensile stress = 76.648 Mpa

2) extension = 0.0215 m

Explanation:

Detailed explanation and calculation is shown in the image below

5 0
3 years ago
What will the stopping distance be for a 2,000-kg car if -2,000 N of force are applied when the car is traveling 20 m/s?
astraxan [27]

Answer is B- 200 m

Given:

m (mass of the car) = 2000 Kg

F = -2000 N

u(initial velocity)= 20 m/s.

v(final velocity)= 0.

Now we know that

<u>F= ma</u>

Where F is the force exerted on the object

m is the mass of the object

a is the acceleration of the object

Substituting the given values

-2000 = 2000 × a

a = -1 m/s∧2

Consider the equation

<u>v=u +at</u>

where v is the initial velocity

u is the initial velocity

a is the acceleration

t is the time

0= 20 -t

t=20 secs


s = ut +1/2(at∧2)

where s is the displacement of the object

u is the initial velocity

t is the time

v is the final velocity

a is the acceleration

s= 20 ×20 +(-1×20×20)/2

<u>s= 200 m</u>


3 0
3 years ago
Read 2 more answers
An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in
Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

T {V}^{\gamma - 1} = \text{constant}

\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}

Given:-

{T}_{1} = 275 \; K  

{T}_{2} = T \left( \text{say} \right)

{V}_{1}  = V

{V}_{2} = 6V

\gamma = \cfrac{5}{3} \;    (the gas is monoatomic)

\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}

 

\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}  

T  =  275 \times 0.30

T  =  83 K.

3 0
3 years ago
Does groundwater flow or stay still?
zhenek [66]

Answer:

yes it flows through flow paths.

Explanation:

5 0
3 years ago
Earth exerts a 100 N gravitational force on a metal box. What is the magnitude of the gravitational force the metal box exerts o
jeka94

We have that the magnitude of the gravitational force  is mathematically given as

f=6.377N

<h3>Force</h3>

Question Parameters:

Earth exerts a 100 N gravitational force on a metal box.

(Mass of the earth is 6e24 kg and radius of the earth is 6.4e6m.)

Generally the equation for the Gravitational mForce  is mathematically given as

F=\frac{GMm}{r^2}\\\\Therefore\\\\F=\frac{100/9.8* 6e116e24}{6.4e6^2}

f=6.377N

For more information on Force visit

brainly.com/question/26115859

7 0
2 years ago
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