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
Korolek [52]
3 years ago
6

How much work does a 65 kg gymnast have to do to climb a rope to a height of 7? meters? PLEASE HURRY ITS A TEST​

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
4 0
How much force did the gymnast use to climb the rope?
You might be interested in
Which of the following units are used to describe acceleration?
Pavel [41]

Answer:

m/s2 m/s

Explanation:

7 0
3 years ago
The table shows the commonly eaten foods of some groups of organisms.
rodikova [14]

Answer: A and B

Explanation:

took the test:) hope this helps

6 0
4 years ago
what is the necessary condition within the hydrogen atom for balmer line photons to be absorbed by the hydrogen atom
luda_lava [24]

Explanation :

When an electron jumps from one energy level to another, the energy of atom gets changed.

If a photon gets absorbed, the electron will move to higher energy levels and then fall back to the lower energy levels. Then each time a photon will be absorbed whose energy is given by difference between the initial and final energy levels i.e

In Balmer series, the transition is from higher energy levels to n = 2.

So, the necessary condition for Balmer series is that the electron should be at first excited state or n = 2 level as shown in figure.


6 0
3 years ago
The drawing shows a large cube (mass = 21.0 kg) being accelerated across a horizontal frictionless surface by a horizontal force
MaRussiya [10]

Answer:

The blocks must be pushed with a force higher than 359 Newtons horizontally in order to accomplish this friction levitation feat.

Explanation:

The first step in resolving any physics problem is to draw the given scenario (if possible), see the attached image to have an idea of the objects and forces involved.

The large cube in red is being pushed from the left by a force \vec{P} whose value is to be found. That cube has its own weight \vec{w}_1=m_1\vec{g}, and it is associated with the force of gravity which points downward. Newton's third law stipulates that the response from the floor is an upward pointing force on the cube, and it's called the normal force \vec{N}_1.

A second cube is being pushed by the first, and since the force \vec{P} is strong enough it is able to keep such block suspended as if it were glued to the first one, due to friction. As in the larger cube, the smaller one has a weight \vec{w}_2=m_2\vec{g} pointing downwards, but the normal force in this block doesn't point upwards since its 'floor' isn't below it, but in its side, therefore the normal force directs it to the right as it is shown in the picture. Normal forces are perpendicular to the surface they contact. The final force is the friction between both cubes, that sets a resistance of one moving parallel the other. In this case, the weight of the block its the force pointing parallel to the contact surface, so the friction opposes that force, and thus points upwards. Friction forces can be set as Fr=\mu~N, where \mu is the coefficient of static friction between the cubes.

Now that all forces involved are identified, the following step is to apply Newton's second law and add all the forces for each block that point in the same line, and set it as equal its mass multiplied by its acceleration. The condition over the smaller box is the relevant one so its the first one to be analyzed.

In the vertical component: \Sigma F^2_y=Fr-w_2=m_2 a_y Since the idea is that it doesn't slips downwards, the vertical acceleration should be set to zero a_y=0, and making explicit the other forces: \mu N_2-m_2g=0\quad\Rightarrow (0.710)N_2-(4.5)(10)=0\quad\Rightarrow N_2=(4.5)(10)/(0.710)\approx 63.38 [N]. In the last equation gravity's acceleration was rounded to 10 [m/s^2].

In its horizontal component: \Sigma F^2_x=N_2=m_2 a_x, this time the horizontal acceleration is not zero, because it is constantly being pushed. However, the value of the normal force and the mass of the block are known, so its horizontal acceleration can be determined: 63.38=(4.5) a_x \quad \Rightarrow a_x=(63.38)/(4.5)\approx 14.08 [m/s^2]. Notice that this acceleration is higher than the one of gravity, and it is understandable since you should be able to push it harder than gravity in order for it to not slip.

Now the attention is switched to the larger cube. The vertical forces are not relevant here, since the normal force balances its weight so that there isn't vertical acceleration. The unknown force comes up in the horizontal forces analysis: \Sigma F_x=P=m a_x, since the force \vec{P} is not only pushing the first block but both, the mass involved in this equation is the combined masses of the blocks, the acceleration is the same for both blocks since they move together; P=(21.0+4.5) 14.08\approx 359.04 [N]. The resulting force is quite high but not impossible to make by a human being, this indicates that this feat of friction suspension is difficult but feasable.

4 0
3 years ago
A sling is used to give a stone an initial velocity of 20 at an angle of 30 above the horizontal. The stone travels through the
Luba_88 [7]

Answer:

Option E is correct.

There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.

Explanation:

Normally, ignoring air resistance, for projectile motion, the range (horizontal distance teavelled) of the motion is given as

R = (u² sin 2θ)/g

where

u = initial velocity of the projectile = 20 m/s

θ = angle above the horizontal at which the projectile was launched = 30°

g = acceleration due to gravity = 9.8 m/s²

R = (30² sin 60°) ÷ 9.8

R = 78.53 m

So, Normally, the stone should travel a horizontal distance of 78.53 m. So, travelling a horizontal distance of 32 m (less than half of what the range should be without air resistance) means that, the motion of the stone was impeded, hence, option E is correct.

There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.

Hope this Helps!!!

7 0
4 years ago
Other questions:
  • To the nearest tenth, what is the area of the shaded segment when JA=8ft ??
    13·2 answers
  • An experiment that produces waves on a string is analyzed, and the experimental value for the wave speed is found to be 66 m/s.
    16·2 answers
  • What does intonation refer to select one of the options below as your answer:a. the change in the pitch (high or low sound of th
    8·1 answer
  • NEED THE ANSWER PLZ<br><br>What are lanthanides?
    5·1 answer
  • You have 1.2kg of (AU)gold (that's quite a bit of money at $1839/ounce). Using a heat source you apply 3096 J to the gold and re
    10·1 answer
  • Example of the center of the gravity<br>​
    14·1 answer
  • The granulation pattern that astronomers have observed on the surface of the Sun tells us that: a. the Sun is a lot cooler on th
    7·1 answer
  • Which line represents a stationary object?<br> Position<br> Time
    7·1 answer
  • What causes sounds that are lower than 0 decibels?
    8·1 answer
  • 5, the following equation shows the position of a particle in time t, x=at2i + btj where t is in second and x is in meter. A=2m/
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!