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
Pepsi [2]
2 years ago
6

Two iron weights, one twice the mass of the other, are dropped from the top of a building. Compared with the lighter weight, the

twice as heavy weight will hit the ground with...
Physics
2 answers:
OverLord2011 [107]2 years ago
8 0
The same acceleration, the same speed, at the same time, with double the momentum, the same potential energy, and double the kinetic energy.
Whitepunk [10]2 years ago
4 0
The twice as heavy weight will hit the ground with more force, or impact.
You might be interested in
Pressure, V is volume , T is temperature of a gas and R is gas constant.
Ilya [14]
هاذغهعفذعذهفذهذهغذهفهذهفذهف
6 0
2 years ago
4. A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exer
Rudik [331]

Answer:

0.61°

Explanation:

Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.

Pulling force= resistance force

From the formula for pulling force,

F(x)= Fcos(θ)

= 425×cos(35.2)

=347N

The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)

=425×0.567=245N

Resistance force= (325N+ 245N) (α)= 570N(α)

We can now equates the pulling force to resistance force

570 (α)= 347N

(α)= 347/570

= 0.61

3 0
3 years ago
1. Indicate whether these objects or atoms are positively,<br> negatively or neutrally charged.
vovangra [49]

Answer:

Neutrally charged!!!!!!!!!!!!!!!!!!!!!

Explanation:

6 0
2 years ago
From a window that is 20 m from the ground a stone with a speed of 10m / s is thrown vertically upwards. Calculate:
Oduvanchick [21]

a)

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the maximum height

v = final velocity at the maximum height = 0 m/s

t = time taken to reach the maximum height

Using the equation

v² = v₀² + 2 a (Y - Y₀)

0² = 10² + 2 (- 9.8) (Y - 20)

Y = 25.1 m


also using the equation

v = v₀ + a t

inserting the values

0 = 10 + (- 9.8) t

t = 1.02 sec


b)

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the ground = 0 m

t = time taken to reach the ground

Using the equation

Y = Y₀ + v₀ t + (0.5) a t²

0 = 20 + 10 t + (0.5) (- 9.8) t²

t = 3.3 sec

3 0
3 years ago
At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There
katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

\tau_{net} = I \alpha (1)

where

\tau_{net} is the net torque acting on the object in rotation

I is the moment of inertia of the object

\alpha is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

\alpha = \frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time interval

So we can rewrite eq.(1) as

\tau_{net}=I\frac{\Delta \omega}{\Delta t}

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

\tau_{net}=0

From the previous equation, this implies that

\Delta \omega =0

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

\tau_{net}=0

Therefore, this means that

\Delta \omega =0

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0
3 years ago
Other questions:
  • A chemical form in which most fat exists in food as well as in the body is known as ______________.
    12·2 answers
  • Two metal balls have charges of 7.1 × 10-6 coulombs and 6.9 × 10-6 coulombs. They are 5.7 × 10-1 meters apart. What is the force
    5·2 answers
  • Do you think the benifits of nuclear power outweigh the potential drawbacks
    11·1 answer
  • Mutations are avoided during replication because DNA polymerase is able to _____?
    13·2 answers
  • Width, height, thickness and thermal conductivity of a single pane window and the air space of a double pane window. Representat
    6·1 answer
  • List 3 objects that would sink in water &amp; 3 object that would float in water :)​
    6·2 answers
  • A river flows at a rate of 2 km divided by h. A patrol boat travels 54 km upriver and returns in a total time of 9 hr. What is t
    9·1 answer
  • At an amusement park, a swimmer uses a water slide to enter the main pool. a. If the swimmer starts at rest, slides with negligi
    8·1 answer
  • A bullet is fired vertically upward a velocity of 80m/s to what height will the bullet rise above the point of projection​,note:
    5·1 answer
  • Base your answer(s) to the following question(s) on the information and diagram
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!