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swat32
3 years ago
11

an aeroplane moving horizontally with a speed of 180 km/hr drops a food packet while flying at a height of 490m . the horizontal

range of the packet is?? ...?
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
5 0
The time the package travels horizontally is equal to the time it takes to hit the ground. This can be calculated using:
s = ut + 1/2 at²; u is 0
480 = 4.9t²
t = 9.90 seconds

Horizontal distance = horizontal speed x time
The speed will be converted to m/s from km/h
= 180 km/hr x 1000m/km x 1hr/3600 seconds x 9.90 seconds
= 495 m
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FIGURE 1 shows part of a mass spectrometer. The whole arrangement is in a vacuum. Negative ions of mass 2.84 x 10-20 kg and char
yuradex [85]

Yes, the ions can exit slit P without being deflected, if the electric field strength is 170.6 N/C

Explanation:

When the ions are inside the container, they are subjected to two forces, with directions opposite to each other:

  • The force due to the electric field, whose magnitude is F_E=qE, where q is the charge of the ion and E is the strength of the electric field
  • The force due to the magnetic field, whose magnitude is F_B=qvB, where v is the speed of the ions and B is the strength of the magnetic field

The ions will move straight and undeflected if the two forces are equal and opposite. By using Fleming Left Hand rule, we notice that the magnetic force on the (negative) ions point upward: this means that the electric field must be also upward (so that the electric force on the ions is downward). Then, the two forces are balanced if

F_E = F_B

which translates into

qE=qvB\\\rightarrow v = \frac{E}{B}

Therefore, if the speed of the ions is equal to this ratio, the ions will go undeflected.

We can even calculate the value of E at which this occurs. In fact, we know that the ions are earlier accelerated by a potential difference V=-3000 V, so we have that their kinetic energy is given by the change in electric potential energy:

qV=\frac{1}{2}mv^2

where

q=-2.0\cdot 10^{-19}C\\m=2.84\cdot 10^{-20}kg

Solving for v, the speed,

v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(-2.0\cdot 10^{-19})(-3000)}{2.84\cdot 10^{-20}}}=205.6 m/s

And since the magnetic field strength is

B = 0.83 T

The strength of the electric field must be

E=vB=(205.6 m/s)(0.83 T)=170.6 N/C

Learn more about electric and magnetic fields:

brainly.com/question/8960054

brainly.com/question/4273177

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

7 0
3 years ago
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
Tim’s cow is anemic. The cow is lacking which type of nutrient?
mixer [17]

Answer:

Iron deficiency

Explanation:

or more scientifically explained as decreased hemoglobin levels in your blood but still caused by lack of iron.

6 0
3 years ago
Read 2 more answers
A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring
sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

4 0
3 years ago
If object A has more mass than object B, what will object A need to accelerate at the same rate as object B?
Leni [432]

Answer:

More force

Explanation:

Object A has more mass than object B

  For object A to accelerate at the same rate as object B, it will need more force.

According to Newton's second law of motion "the net force on a body is the product of its mass and acceleration".

  Net force  = mass x acceleration

Now, if a body has more mass and needs to accelerate at the same rate as another one with a lower mass, the force on it must be increased.

3 0
2 years ago
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