What two aspects of a force do scientists measure?

I NEED ANSWERS!

Neporo4naja [7]

Answer:

A water fall?

Explanation:

It's an example I took in my science class

5 0

3 years ago

Read 2 more answers

a measuring cylinder contains 30cm3 of a liquid some more of the liquid is added until the liquid level reaches the 50cm3 mark t

S_A_V [24]

Answer:

1.5g/cm³

Explanation:

Given parameters:

Initial volume of the liquid = 30cm³

Final volume of the liquid = 50cm³

Mass of the liquid = 30g

Unknown:

Density of the liquid = ?

Solution:

Density is the mass per unit volume of a substance.

Density = $\frac{mass}{volume}$

volume of the liquid = final volume - initial volume = 50cm³ - 30cm³

volume of the liquid = 20cm³

Density = $\frac{30}{20}$ = 1.5g/cm³

6 0

3 years ago

10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr

12345 [234]

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

$F_{g0}=G \frac{mM_{E0}}{r^{2}}$

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=2M_{E0}$

so:

$F_{gf}=G \frac{2mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=2$

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=3M_{E0}$

so:

$F_{gf}=G \frac{3mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=3$

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

$M_{Ef}=\frac{M_{E0}}{4}$

so:

$F_{gf}=G \frac{mM_{E0}}{4r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=\frac{1}{4}$

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

4 0

3 years ago

what is the initial velocity of a go-kart traveling at a uniform acceleration of 0.5 m/s^2 for 5s as it slows down to a stop?

timurjin [86]

The initial velocity of go-kart is 2.5 m/s.

<u>Explanation:</u>

Here, the uniform acceleration of go-kart is given as 0.5 m/s². Also the time required by it to stop is also given as 5 s. As acceleration is the measure of change in velocity per unit time.

In this case, the velocity should be changed from a value to zero to come to rest. So the initial velocity will be positive value and final velocity is zero.

As we know the values of acceleration, final velocity and time, the initial velocity can be easily determined as follows.

$Acceleration = \frac{Final velocity -Initial velocity}{Time}$

Since, final velocity is zero, acceleration is 0.5 m/s² and time is 5 s, then,

$-0.5=\frac{-\text {Initial velocity}}{5}$

Initial velocity = 0.5 × 5 = 2.5 m/s.

So the initial velocity of go-kart is 2.5 m/s.

8 0

3 years ago

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.45 m/s^2 for 20.0 s.

Anton [14]

Answer:

a) The total displacement of the trip was 5.32 × 10³ m

b) The average speeds were:

leg 1: 24.5 m/s

leg 2: 49 m/s

leg 3: 23.9 m/s

Complete trip: 43.8 m/s

Explanation:

The position and velocity equations for an object moving along a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

If the velocity is constant, then a = 0 and x = x0 + v · t where "v" is the velocity.

a) To calculate the total displacement of the trip, let´s calculate the distance traveled in each phase.

Phase 1:

x = x0 + v0 · t + 1/2 · a · t²

x = 0 m + 0 m/s · t + 1/2 · 2.45 m/s² · (20.0 s)²

x = 490 m

The velocity reached in that phase is:

v = v0 + a · t

v = 0 m/s + 2.45 m/s² · 20.0 s

v = 49.0 m/s

Phase 2:

x = x0 + v · t

x = 490 m + 49.0 m/s · 96.0 s

x = 5.19 × 10³ m

Phase 3:

x = x0 + v0 · t + 1/2 · a · t²

x = 5.19 × 10³ m + 49 m/s · 5.44 s - 1/2 · 9.00 m/s² · (5.44 s)²

x = 5.32 × 10³ m

The total displacement of the trip was 5.32 × 10³ m

b) The average speed is calculated as the traveled distance divided by the elapsed time:

average speed v = final position - initial position / (final time- initial time)

Phase 1:

v = 490 m - 0 m / 20.0 s = 24.5 m/s

Phase 2:

v = 5.19 × 10³ m - 490 / 96.0 s

v = 48.9 m/s (without rounding the final position the result is 49.0 m/s)

Phase 3:

v = 5.32 × 10³ m - 5.19 × 10³ m / 5.44 s = 23.9 m/s

For the complete trip:

v = 5.32 × 10³ m - 0 m / (20.0 s + 96.0 s + 5.44 s)

v = 43.8 m/s

7 0

4 years ago