Why is it good for scientist to be skeptical

If you push a 900N refrigerator with the same force as you lift a 40N dumbbell, the dumbbell will move faster. Which of Newton’s

svetlana [45]

This is Newton's second law of motion.

It states that Force exerted on a moving object is directly proportional to its product of mass and acceleration.

i.e Force(F) = Mass(M) × Acceleration(A)

Because Force is constant here, reducing the weight of the body the force is applied to consequently reduces its mass.

If mass is reduced, the body's acceleration increases hence increasing its speed.

That is why the 40N dumbbell move faster.

3 0

3 years ago

Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of 2

Sidana [21]

Answer:

The work done on the wagon is 37 joules.

Explanation:

Given that,

The force applied by Charlie to the right, F = 37.2 N

The force applied by Sara to the left, F' = 22.4 N

We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :

$F_n=F-F'$

$F_n=37.2-22.4$

$F_n=14.8\ N$

Work done on the wagon is given by the product of net force and displacement. It is given by :

$W=F_n\times d$

$W=14.8\ N\times 2.5\ m$

W = 37 Joules

So, the work done on the wagon is 37 joules. Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers

A roller coaster at the Six Flags Great America amusement park in Gurnee, Illinois, incorporates some clever design technology a

sergeinik [125]

Answer:

a)radius of the arc of the teardrop at the top is 10.58m

b)T = mg

c) the centipetal acceleration at the top is 14.5m/s

Explanation:

Part A

Given that roller coaster is of tear drop shape

So the speed at the top is given as

v = 14.4 m/s

acceleration at the top is given as

a = 2 g

$a = 2(9.8) = 19.6 m/s^2$

now we know the formula of centripetal acceleration as

$a = \frac{v^2}{R}\\$

$19.6 = \frac{14.4^2}{R}\\R = 10.58 m$

Part B

now

the total mass of the car and the ride is M

Let the force exerted by the track be n

By Newton law

$n +Mg =\frac{Mv^2}{r} \\\\n=\frac{Mv^2}{r} -Mg\\\\=M(\frac{v^2}{r}-g )\\\\=M(2g-g)\\\\T=Mg$

Part C

If the radius of the loop is 21.4 m

speed is given by same v = 14.4 m/s

now the acceleration is given as

$a = \frac{v^2}{R}$

$a = \frac{14.4^2}{21.4} \\\\= 9.69 m/s^2$

Now for normal force at the top is given by force equation

$F_n + mg = ma\\F_n = m(a-g)$

The force exerted by the rail is less than zero because acceleration is less than 9.69m/s²

So the normal force would have to point away from the centre, For safe ride this normal force must be positive i.e $a \prec g$

$\frac{v^2}{r} \prec \sqrt{g} \\\\v = \sqrt{rg} \\\\v = \sqrt{21.4 \times 9.8} \\\\v = 14.5m/s$

3 0

3 years ago

A bucket of water with mass 5 kg sits on the ground with a coefficient of static friction of 0.35. What is the maximum force of

allochka39001 [22]

Answer:

The force of static friction is 17.15 N

Explanation:

It is given that,

Mass of the bucket, m = 5 kg

The coefficient of static friction is, $\mu=0.35$

We need to find the maximum force of static friction. It is given by :

$F=\mu mg$

$F=0.35\times 5\ kg\times 9.8\ m/s^2$

F = 17.15 N

So, the force of static friction is 17.15 N. Hence, this is the required solution.

6 0

4 years ago

Use the graph of velocity versus time for an object to answer the question.

Pavel [41]

T<u>he statement which fairly compares segment 2 and segment 3 </u>is These represent equal periods of time, but the force during segment 2 is different than the force during segment 3.

Since segment 2 starts at t = 60 s and ends at t = 150 s, the time interval is Δt = 150 - 60 = 90 s.

Also, segment 3 starts at t = 150 s and ends at t = 240 s, the time interval is Δt = 240 - 150 = 90 s.

So, their time periods are the equal.

We notice that segment 2 is less steep than segment 3 this implies that the acceleration in each segment is different, since the acceleration is the slope of the graph.

Since force is determined by acceleration, this implies that the force on segment 2 is different form the force acting in segment 3.

So, we have equal time periods but different forces.

So, <u>the statement which fairly compares segment 2 and segment 3 </u>is These represent equal periods of time, but the force during segment 2 is different than the force during segment 3.

Learn more about velocity-time graph here:

brainly.com/question/24788847

8 0

3 years ago