1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
padilas [110]
3 years ago
11

You push really hard against a round rock, but you cannot make it roll. Which statement best explains why you cannot move the ro

ck?
Gravity pulls the rock toward the center of Earth.
The forces between you and the rock are balanced.
The force of your push is greater than the opposing force.
There is no friction to help you move the object.
Physics
1 answer:
Burka [1]3 years ago
8 0

Answer:

Its B

Explanation:

I searched it up :>

You might be interested in
a 0.0215m diameter coin rolls up a 20 degree inclined plane. the coin starts with an initial angular speed of 55.2rad/s and roll
marissa [1.9K]

Answer:

h = 0.0362\,m

Explanation:

Given the absence of non-conservative force, the motion of the coin is modelled after the Principle of Energy Conservation solely.

U_{g,A} + K_{A} = U_{g,B} + K_{B}

U_{g,B} - U_{g,A} = K_{A} - K_{B}

m\cdot g \cdot h = \frac{1}{2}\cdot I \cdot \omega_{o}^{2}

The moment of inertia of the coin is:

I = \frac{1}{2}\cdot m \cdot r^{2}

After some algebraic handling, an expression for the maximum vertical height is derived:

m\cdot g \cdot h = \frac{1}{4}\cdot m \cdot r^{2}\cdot \omega_{o}^{2}

h = \frac{r^{2}\cdot \omega_{o}^{2}}{g}

h = \frac{(0.0108\,m)^{2}\cdot (55.2\,\frac{rad}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }

h = 0.0362\,m

3 0
3 years ago
A ball of mass 2kg falls from a height of 200m. Calculate it's kinetic energy after falling a distance of 50m . (g=10m/s²) . Neg
SCORPION-xisa [38]

Answer:

U = 1000 J

Explanation:

By the conservation of energy, potential energy is converted into kinetic energy when the ball falls from a height.

 U = \frac{1}{2} mv^2 = mgh

U = Energy (in Joules)

m = mass (in kg)

v = velocity in (m/s)

g = acceleration due to gravity (in m/s^2)

h = height (in m)

Let's use mgh as we have the values given.

U = 2(10)(50)

U = 1000 J

4 0
3 years ago
Suppose a lion running at a velocity of 10 m/s east slows
iris [78.8K]

Answer:

a=0.34\ m/s^2

Explanation:

Given that,

Initial velocity, u = 10 m/s

Final velocity, v = 0 (stops)

Time, t = 30 s

We need to find the acceleration of the lion.

We know that,

Acceleration = rate of change of velocity

a=\dfrac{v-u}{t}\\\\a=\dfrac{10-0}{30}\\\\a=0.34\ m/s^2

So, the acceleration of the lion is 0.34\ m/s^2.

5 0
3 years ago
A big olive (* - 0.50 kg) lies at the origin of an xy coordinate system, and a big BrazlI nut (M - 1.5^kg) lie^s at the point (1
Afina-wow [57]

The <em>estimated</em> displacement of the center of mass of the olive is \overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m].

<h3>Procedure - Estimation of the displacement of the center of mass of the olive</h3>

In this question we should apply the definition of center of mass and difference between the coordinates for <em>dynamic</em> (\vec r) and <em>static</em> conditions (\vec r_{o}) to estimate the displacement of the center of mass of the olive (\overrightarrow{\Delta r}):

\vec r - \vec r_{o} = \left[\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot(m_{i}\cdot g + F_{i, x})}{\Sigma \limits_{i =1}^{2}(F_{i,x}+m_{i}\cdot g)} ,\frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot(m_{i}\cdot g + F_{i, y})}{\Sigma \limits_{i =1}^{2}(F_{i,y}+m_{i}\cdot g)} \right]-\left(\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}, \frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}\right) (1)

Where:

  • r_{i, x} - x-Coordinate of the i-th element of the system, in meters.
  • r_{i,y} - y-Coordinate of the i-th element of the system, in meters.
  • F_{i,x} - x-Component of the net force applied on the i-th element, in newtons.
  • F_{i,y} - y-Component of the net force applied on the i-th element, in newtons.
  • m_{i} - Mass of the i-th element, in kilograms.
  • g - Gravitational acceleration, in meters per square second.

If we know that \vec r_{1} = (0, 0)\,[m], \vec r_{2} = (1, 2)\,[m], \vec F_{1} = (0, 3)\,[N], \vec F_{2} = (-3, -2)\,[N], m_{1} = 0.50\,kg, m_{2}  = 1.50\,kg and g = 9.807\,\frac{kg}{s^{2}}, then the displacement of the center of mass of the olive is:

<h3>Dynamic condition\vec{r} = \left[\frac{(0)\cdot (0.50)\cdot (9.807)+(0)\cdot (0) + (1)\cdot (1.50)\cdot (9.807) + (1)\cdot (-3)}{(0.50)\cdot (9.807) + 0 + (1.50)\cdot (9.807)+(-3)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (0)\cdot (3) + (2)\cdot (1.50)\cdot (9.807) +(2) \cdot (-2)}{(0.50)\cdot (9.807) + (3)+(1.50)\cdot (9.807)+(-2)}  \right]\vec r = (0,704, 1.233)\,[m]</h3>

<h3>Static condition</h3><h3>\vec{r}_{o} = \left[\frac{(0)\cdot (0.50)\cdot (9.807) + (1)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807) + (1.50)\cdot (9.807)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (2)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807)+(1.50)\cdot (9.807)}  \right]</h3><h3>\vec r_{o} = \left(0.75, 1.50)\,[m]</h3><h3 /><h3>Displacement of the center of mass of the olive</h3>

\overrightarrow{\Delta r} = \vec r - \vec r_{o}

\overrightarrow{\Delta r} = (0.704-0.75, 1.233-1.50)\,[m]

\overrightarrow{\Delta r} = (-0.046, -0.267)\,[m]

The <em>estimated</em> displacement of the center of mass of the olive is \overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]. \blacksquare

To learn more on center of mass, we kindly invite to check this verified question: brainly.com/question/8662931

3 0
2 years ago
A steam stream of 400 m3 /s, standard (standard cubic meter per second) is cooled by adding 7 m3 /s cold water. The fluid leavin
antiseptic1488 [7]

Answer: 510 m/s

Explanation: specific gravity of steam is 18/29 = 0.620

It is the ratio of the density of steam over density of water

400m3/s of steam =

400m3ms * 0.620 of water

= 248m3/s of water

Total flow rate Q = 248 + 7 = 255m3/s

Using Q = AV

Where A is area and V is velocity

V = Q/A

V = 255/0.5 = 510m/s

4 0
3 years ago
Other questions:
  • A man tries to push a box of mass 4 kg with a force of 5 N but the box remains stationary If the coefficient of static friction
    13·1 answer
  • A cue ball has a mass of 0.5 kg. During a game of pool, the cue ball is struck and now has a velocity of 3 . When it strikes a s
    13·2 answers
  • Which form of carbon in the figure above is the hardest
    6·1 answer
  • A cat of mass of 1kg runs at a speed of 2.5 m/s what is the momentum of the cat?
    5·2 answers
  • An airplane flying horizontally at a constant speed of 350 km/h over level ground releases a bundle of food supplies. Ignore the
    9·1 answer
  • When the radius of curvature of the lens increases, what happens to the focal length of the lens
    5·1 answer
  • What are the elements needed for effective teamwork?
    14·2 answers
  • The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
    10·1 answer
  • Electromagnetic waves are
    11·2 answers
  • Which of these talents is an HR employee MOST likely to need?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!