A wheelbarrow is experiencing two forces. One is pushing 5 N to the east and the other 5 N in the exact opposite direction. What must be true of the wheelbarrow's motion?

Answers

Answer 1

Answer:

Answer:

its acceleration must be 0.

Explanation:

According the second Newton's law, the net force applied in a object is equal to the mass of the object multiplied by its acceleration:

$F_(net) =m.a$

In this case, the net force is

$F_(net)=5N-5N=0$

What makes sense since both forces have the same value and opposite directions.

Since the mass of the object can't be zero, the acceleration is the one that must be 0 in order fulfill the equation of the second newton's law.

Another option is that, since both forces are opposite, the wheelbarrow could be at rest. Though this is correct, this is not always true.

In conclusion, the only statement that always is true about the wheelbarrow's motion is that its acceleration must be 0.

Answer 2

Answer: Assuming no other forces are acting on the wheelbarrow, it must be stationary.

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Explain the energy transformations that occur when breaking in a hybrid
vehicle.

Answers

it occur kinetic energy

when breaking in a hybrid

vehicle.

hope it helps

Answer:

it occur kinetic energy when breaking in a hybrid

vehicle .

hope it is helpful to you ☺️

In a motor, the combined effect of electric currents and magnetic forces turn electrical energy into which of the following? A.
a battery

B.
a generator

C.
mechanical energy

D.
electromagnetic energy

Need help
@Akin
@AL2006

Answers

it doesn't turn it into a battery (storatge of energy) or a generator (maker of energy)

it is making it into a type of energy

mechanical or electromagnetic energy

if you look at how a motor works, the electric current passes throught the wire and alternates csueing the motor to spin so I would say electormagnetic

Answer:

C. Mechanical Energy

Explanation:

Took the test

Why do prokaryotic cells usually reproduce more quickly than eukaryotic cells? Choose 1 answer:

(Choice A) A Eukaryotic cells are more structurally complex than prokaryotic cells.

(Choice B) B Eukaryotic cells have less DNA than prokaryotic cells.

(Choice C) C Eukaryotic cells have more cell walls than prokaryotic cells.

(Choice D) D Eukaryotic cells are smaller than prokaryotic cells.

Answers

Prokaryotic cells reproduce more quickly than eukaryotic cells because the latter are more structurally complex. The correct option is A.

Structural complexity of cells

Eukaryotic cells have nuclei and membrane-bound organelles such as mitochondrion. This is unlike prokaryotic cells whose DNA lies freely in the cytoplasm and lacks membrane-bound organelles.

In order words, eukaryotic cells are structurally more complex than prokaryotic cells. Thus, reproductive times in prokaryotes are usually quicker than in eukaryotes.

More on the structural complexity of cells can be found here: brainly.com/question/3502696

#SPJ1

A marble is thrown horizontally with a speed of 15.0 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 65.0° with the horizontal. From what height above the ground the marble was thrown. How far from the base of the building, did the marble land?

Answers

The answer is 52.79 m.

I used this formula to get the formula for Vy:
v^2=vi^2+2(a)(x)
And got
Vy=square root (19.6 h)

Then I used that and put it in this formula:

tan(65) =Vy/Vx

tan(65) = square root (19.6 h)/15.0

Then I rearranged it to:

h=[(15.0)(tan65)]/19.6

h=52.79 m

The answer for this question is 52.79

A sample of gold has a mass of 38.6 grams and a volume of 2 cm3. What is the density of gold

Answers

The density of the sample of gold whose mass is 38.6 grams and with a volume of 2 cm³ is 19300 kg/m

Density: This can be defined as the ratio of mass and velocity. The S.I unit of density is kg/m³.

The question above can be solved using the formula below

D = m/V.................... Equation 1

Where D = density of the sample of gold, m = mass of gold, V = volume of gold.

From the question,

Given: m = 38.6 g, = 0.0386 kg, V = 2 cm³ = 0.000002 m³

Substitute these values into equation 1

D = 0.0386/0.000002

D = 19300 kg/m³

Hence the density of the sample of gold is 19300 kg/m³

Learn more about density here: brainly.com/question/23562510

Answer:

$19.3 g/cm^3$

Explanation:

The density of a material is given by

$\rho = (m)/(V)$

where

m is the mass of the sample

V is its volume

for the sample of gold in this problem,

m = 38.6 g

$V=2 cm^3$

Substituting into the formula, we find the density of gold:

$\rho = (38.6)/(2)=19.3 g/cm^3$

What is the minimum speed with which he’d need to run off the edge of the cliff to make it safely to the far side of the river? the answer was 6 m/s the world-record time for the 100 m dash is approximately 10 s. given this, is it reasonable to expect brady to be able to run fast enough to achieve brady's leap? a.yes, the obtained speed is less than the world-record.
b.yes, the obtained speed is almost equal to the world-record.
c.no, the obtained speed is greater than the world-record.
d.no, the obtained speed is almost equal to the world-record.

Answers

The answer is

A. Yes, the obtained speed is less than the world record

The explanation:

when the obtained speed is 6 m /s

and the world record speed = distance / time = 100 m / 10 s = 10 m/s

So, Yes, the obtained speed is less than the world record

Not only is it less, its also a reasonable average speed for a somewhat athletic person. Therefore, the leap is entirely possible.

-Samuel Brady gained his lasting notoriety for his leap over the Cuyahoga River around 1780 in what is now Kent, Ohio. After following a band of Indians into the Ohio country, a failed ambush attempt resulted in the band chasing Brady near the Cuyahoga River. To avoid capture, Brady leaped across a 22-foot (6.7 m) wide gorge of the river (which was widened considerably in the 1830s for construction of the Pennsylvania and Ohio Canal) and fled to a nearby lake where he hid in the water under a fallen tree using a reed for air.

Part (a): The minimum speed of Brady should be $\boxed{6\text{ m/s}}$ to cross the cliff.

Part (b): The speed of Brady’s leap is possible to achieve, as it is less than the speed of world record.

Further Explanation:

(a)

Brady jumps from a cliff to go on other side. He crosses $22\text{ ft}$ horizontal distance from a height of $20\text{ft}$ . To jump from cliff, he follows the Newton’s law of motion.

Given:

The horizontal distance is $22\text{ ft}$ .

The vertical distance is $20\text{ft}$ .

Concept:

The horizontal distance in meter is $6.71\text{ m}$ .

The vertical distance in meter is $6.10\text{ m}$ .

To obtain the time of flight, applying one of the equation of motion given as:

$s=ut+(1)/(2)at^2$

Substitute $0\text {m/s}$ for $u$ and rearrange the above equation for $t$ :

$t=\sqrt{(2s)/(a) }$

Substitute $6.10\text{ m}$ for $s$ and $9.81\text{ m}/\text{s}^2$ for $a$ in above equation.

$\begin{aligned}t&=\sqrt{(2*6.10)/(9.81)}\text{ s}\n&=1.12\text{ s}\end{aligned}$

The time of flight and time taken to cover horizontal distance are equal.

Horizontal velocity of Brady given as:

$\begin{aligned}v&=(6.71)/(1.12)\text{ m/s}\n&\approx6\text{ m/s}\end{aligned}$

Thus, the minimum speed of Brady should be $\boxed{6\text{ m/s}}$ to cross the cliff.

(b)

The minimum speed with which Brady is running is $6\text{ m/s}$ .

The speed of world record given as:

$\begin{aligned}V&=(100)/(10)\text{ m/s}\n&=10\text{ m/s}\end{aligned}$

Thus, the speed of Brady’s leap is possible to achieve, as it is less than the speed of world record.

Learn more:

1. Projectile motion of a body: brainly.com/question/11023695

2. Body in pure rolling motion: brainly.com/question/9575487

3. Newton’s law of motion: brainly.com/question/6125929

Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

1780, Brady's Leap, Captain, Sam, U.S. Continental Army, horizontally, cliff, Ohio's Cuyahoga, gorge, leap, 22 ft, 20 ft, minimum, speed, river, 100 m, dash, 10 s, jump, time of flight, vertical and distance.

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