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alexdok [17]
3 years ago
13

A net force of 10 N accelerates an object at 5.0 m/s^2. What is the mass of the object? A. 2 kg B. 10 kg C. 5 kg D. 20 kg

Physics
1 answer:
Ludmilka [50]3 years ago
7 0

Answer:

A. 2 kg

Explanation:

Force = 10N\\Acceleration = 5.0m/s^2\\Mass = ? = x\\Force = mass\times acceleration \\10 = x \times 5\\10 = 5x \\\frac{10}{5} =\frac{5x}{5} \\2=x\\\\Mass= 2kg

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A 1.0 kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harm
Marina86 [1]

Answer:

v(0)=2m/s

Explanation:

The instantaneous velocity of a point mass that executes a simple harmonic movement is given by:

v(t)=\omega  *A*cos(\omega t + \phi)

Where:

\omega=Angular\hspace{3}frequency\\A=Amplitude\\\phi=Initial\hspace{3}phase

Express the amplitude in meters:

10cm*\frac{1m}{100cm} =0.1m

The angular frequency can be found using the next equation:

\omega=\sqrt{\frac{k}{m} }

Using the data provided:

\omega=\sqrt{\frac{400}{1} } =20

At the equilibrium position:

\phi=0

v(0)=20*(0.1)cos(20*0+0)=2*cos(0)=2*1=2m/s

6 0
3 years ago
Which image represents the force on a positively charged particle caused by an approaching magnet?
Amanda [17]

Answer:

Image B represents the force on a positively charged particle caused by an approaching magnet.

Explanation:

The most fundamental law of magnetism is that like shafts repulse each other and dissimilar to posts pull in one another; this can without much of a stretch be seen by endeavoring to put like posts of two magnets together. Further attractive impacts additionally exist. On the off chance that a bar magnet is cut into two pieces, the pieces become singular magnets with inverse shafts. Also, pounding, warming or winding of the magnets can demagnetize them, on the grounds that such dealing with separates the direct game plan of the particles. A last law of magnetism alludes to maintenance; a long bar magnet will hold its magnetism longer than a short bar magnet. The domain theory of magnetism expresses that every single enormous magnet involve littler attractive districts, or domains. The attractive character of domains originates from the nearness of significantly littler units, called dipoles. Iotas are masterminded in such a manner in many materials that the attractive direction of one electron counteracts the direction of another; in any case, ferromagnetic substances, for example, iron are unique. The nuclear cosmetics of these substances is with the end goal that littler gatherings of particles unite as one into zones called domains; in these, all the electrons have the equivalent attractive direction.

4 0
2 years ago
Can someone explain what is loss of seismic energy ?
xxTIMURxx [149]

Answer:

Seismic attenuation describes the energy loss experienced by seismic waves as they propagate. It is controlled by the temperature, composition, melt content, and volatile content of the rocks through which the waves travel.

Explanation:

5 0
3 years ago
A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
3 years ago
The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. What is the vertical component of t
Lyrx [107]

Given that

Velocity of missile (v) = 20 m/s ,

Angle of missile (Θ) = 53°

Determine ,   Vertical component  = v sin Θ

                                                        = 20 sin 53°

                                                        = 15.97 m/s

6 0
3 years ago
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