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
denis-greek [22]
3 years ago
13

a box is pulled by by two forces. the first force is 50 newtons west and the second force is 20 newtons east. find the resultant

force

Physics
1 answer:
Oliga [24]3 years ago
8 0
Hope this helps you!

You might be interested in
The water hyacinth was a major weed in Australian
Elanso [62]
Water hyacinth can fully cover lakes and wetlands, displacing native aquatic species, reducing fish oxygen levels, and providing optimal habitat for disease-carrying mosquitoes. Water hyacinth infestations can obstruct river transportation, fishing, damage bridges, and clog dams.
6 0
3 years ago
A scene in a movie has a stuntman falling through a floor onto a bed in the room below. The plan is to have the actor fall on hi
tekilochka [14]

Answer:

The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg

Explanation:

Hi there!

Due to conservation of energy, the potential energy (PE) of the mass at a height of 3.32 m will be transformed into elastic potential energy (EPE) when it falls on the mattress:

PE = EPE

m · g · h = 1/2 k · x²

Where:

m = mass.

g = acceleration due to gravity.

h = height.

k = spring constant.

x = compression distance

The maximum compression distance is 0.1289 m, then, the maximum elastic potential energy will be the following:

EPE =1/2 k · x²

EPE = 1/2 · 65144 N/m · (0.1289 m)² = 541.2 J

Then, using the equation of gravitational potential energy:

PE = m · g · h =  541.2 J

m =  541.2 J/ g · h

m = 541.2 kg · m²/s² / (9.8 m/s² · 3.32 m)

m = 16.6 kg

The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg.

6 0
3 years ago
initially, a particle is moving at 5.33 m/s at an angle of 37.9° above the horizontal. Two seconds later, its velocity is 6.11 m
Ainat [17]

Explanation:

Average acceleration is the change in velocity over the change in time:

a = (v − v₀) / t

In the x direction:

aₓ = (6.11 cos (-54.2°) − 5.33 cos (37.9°)) / 2.00

aₓ = -0.316 m/s²

In the y direction:

aᵧ = (6.11 sin (-54.2°) − 5.33 sin (37.9°)) / 2.00

aᵧ = -4.11 m/s²

7 0
3 years ago
5 The frequencies of AM radio waves,
aleksklad [387]
Message me back and I can help you.
5 0
4 years ago
Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24
Oksi-84 [34.3K]
Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}
Now At xp we have:
\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0
\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0
Which is a second order equation, using the quadratic formula to solve for xp would give us:
xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}
or
xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

6 0
3 years ago
Other questions:
  • A rod made up of certaon metall which had coefficiebt of volume expansion is 33×10^-6 per celsius degrer.The rod is 100 meter lo
    5·1 answer
  • You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel wi
    9·1 answer
  • For question #17, use the following picture:
    9·1 answer
  • How can you identify a moveable pulley?
    7·1 answer
  • An electron moves at a speed of 1000 m/s perpendicular to the direction of a uniform magnetic field of 0.5 T. What is the radius
    9·2 answers
  • If the object moves right, which of the following is a possible value of F?
    12·1 answer
  • A soap bubble was slowly enlarged from a radius of 4cm to 6cm. The amount of work necessary for enlargement was 1.5 x 10^-4 joul
    7·1 answer
  • Which electromagnetic wave can be used to treat cancer? Group of answer choices a.microwaves b.gamma waves c.infrared waves d.xr
    6·1 answer
  • I need a answer asap!​
    7·1 answer
  • Exercising at your highest peak
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!