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Over [174]
1 year ago
15

Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball

to hit the ground.(a)Define your coordinate system, be thorough.(b)Write down the given infor-mation, be sure to include hidden information.(c)State what physics principleis at play here. How do you know this?(d)Select an equation.(e)Solve forthe time it takes for the ball to hit the ground.
Physics
1 answer:
Nimfa-mama [501]1 year ago
6 0

Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

b) in this system the equations of motion are

              y = v₀ t + ½ g t²

where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth

e)    y = 0 + ½ g t²

     t = √ (2y / g)

     t = √(2 125 / 9.8)

     t = 5.05 s

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The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from
Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

7 0
1 year ago
In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50μC and initially moving north
Goshia [24]

Answer:

a) The sign of the charge is positive.

b) The magnetic force on the particle is 0.050 newtons.

Explanation:

The magnetic force F on a moving charge with velocity v passing through a magnetic field B is:

\overrightarrow{F}=q\overrightarrow{v}\times\overrightarrow{B}(1)

a)

Because it is a cross product, we can find the direction of the force using the right-hand rule, that is too the direction of the movement. We have two possibilities here because the velocity vector and magnetic field are perpendicular: the particle deflects towards east or toward west, which depends on the charge of the particle. Note that if you put your right hand fingers, except thumb, pointing towards north (direction of velocity) and later close them in the direction of the magnetic field, if you maintain your thumb perpendicular to this movement it will point towards east (See figure), so that will be de direction of the force if the charge is positive, but if the charge is negative, the direction will be opposite (towards west). So the charge has to be positive to deflects towards east.

b)

Now by 1:

F=qvB\sin\theta=(8.50\times10^{-6})(4750)(1.25)\sin90\simeq\mathbf{0.050\,N}

5 0
2 years ago
The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy o
MrMuchimi

Answer:

The total kinetic energy of both particles is 2.43\times10^{-13}

Explanation:

Given that,

Kinetic energy of nucleusK.E= 1.24\times10^{-13}\ J

Kinetic energy of proton K.E= 2.47\times10^{-13}\ J

Radius of proton r= 0.9\times10^{-15}\ m

We need to calculate the final potential energy

Using formula of final potential energy

U=\dfrac{kq^2}{r}

Put the value into the formula

U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times0.9\times10^{-15}}

U_{f}=1.28\times10^{-13}\ J

We need to calculate the initial energy of both the particles

Using formula of energy

E_{i}=(K.E_{n}+K.E_{p})+U_{i}

E_{i}=1.24\times10^{-13}+2.47\times10^{-13}+0

E_{i}=3.71\times10^{-13}\ J

We need to calculate the total kinetic energy of both particles

Using conservation of energy

E_{i}=E_{f}

E_{i}=K.E_{f}+U_{f}

3.71\times10^{-13}=K.E_{f}+1.28\times10^{-13}

K.E_{f}=3.71\times10^{-13}-1.28\times10^{-13}

K.E_{f}=2.43\times10^{-13}

Hence, The total kinetic energy of both particles is 2.43\times10^{-13}

3 0
1 year ago
What is a pesticide?
Brrunno [24]

A pesticide is something that is used to kill / deter pests that eat / destroy crop.

4 0
1 year ago
Read 2 more answers
Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu
Anit [1.1K]
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>

m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... y direction

Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 =  625N^2 =>

Net force = √625 = 25N

F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2

Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. </span>

<span>m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N


Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2

Answer: 6.08 m/s^2


</span>
8 0
1 year ago
Read 2 more answers
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