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3 years ago

5

Suppose a baseball and a marble are dropped att the same time from the same height. which ball would land first according to ari

stotle explain

1 answer:

Natasha_Volkova [10]3 years ago

3 0

The baseball would fall faster, because it has more mass

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(a) What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) For a hydrogen atom, determine the

crimeas [40]

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

$E=\frac {-13.6}{n^2}\ eV$

(a) The energy of the electron in n= 6 excited state is:

$E=\frac {-13.6}{6^2}\ eV$

$E=-0.3778\ eV$

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

$E=\frac {-13.6}{1^2}\ eV$

$E=-13.6\ eV$

$Ratio=\frac {E_6}{E_1}$

$Ratio=\frac {-0.3778}{-13.6}$

Ratio = 0.0278

7 0

4 years ago

A train slows from 60m/s to 20m/s in 50s

Dovator [93]

Answer:

a = -4/5 m/s^2

Explanation:

Acceleration = change in velocity / time

change in velocity = final velocity - initial velocity

a = (20 m/s - 60 m/s) / 50 s

a = -40 m/s / 50 s

a = -4/5 m/s^2

hope this helps! <3

7 0

1 year ago

Read 2 more answers

A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld

lesantik [10]

Answer: The magnitude of the proton's acceleration is 0.748 ×10^14 m/s²

Explanation:
the velocity ,v, of ththe proton = 5.02×10^6 m/s
Magnitude , B , of the magnetic field = 0.180 T

First , we need to find the magnitude of the Force on the proton. This is given by the relation :
F = q(v x B) = qvBsinθ

where 'q' is the charge of proton , q= 1.6×10^-19 C
θ is the angle the proton makes with the direction of the magnetic field

Putting the respective values of v, B ,θ in the above equation, we get:

F = (1.6×10^-19 C)(5.02×10^6 m/s)(0.180T) sin60°
∴ F = 1.25 ×10^-13 N

Now , from Newton's second law we know that ,
F=m×a

∴ a = F/m

Mass of a proton = 1.67×10^27 kg
a= 1.25 × 10^-13 N / 1.67 × 10^27 kg

a= 0.748 × 10^14 m/s² =acceleration of the proton

(To know more about Magnetic Fields : brainly.com/question/9095546)

5 0

2 years ago

After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the driver, with an average

belka [17]

We will define the Total mass to calculate the force, so our values are: $M_p = 69*27=1823Kg\\M_g=15*3=45Kg\\M_c=3*5=15Kg\\M_B=25Kg$

Total Mass $= 1863+45+15+25=1948Kg$

The Weight is,

$F=mg=1948*98=19090.4N$

Through the hook's Law we calculate X.

$F_s=Kx$ , where x is the lenght of compression and K the Spring constant.

We don't have a K-Spring, but we can assume a random value (or simply let the equation in function of K)

$X = \frac{F_s}{x} \\X = \frac{1909.4}{k}$

I assume a value of $K=4*10^4N/m$

$X= \frac{1909.4}{4*10^4} = 0.48m$

6 0

3 years ago

What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma

Sindrei [870]

Answer:

E = 1.50 × $10^{8}$ V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

$\mu _b = \frac{B}{2\mu _o}$ ...............1

and

energy density due to electric filed is

$\mu _e = \frac{\epsilon _o E^2}{2}$ ...............2

and here $\mu _b = \mu _ e$

so that

E = $\frac{B}{\sqrt{\mu _o \times \epsilon _o}}$ ...................3

put here value and we get

$E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}$

E = 3 × $10^{8}$ × 0.50

E = 1.50 × $10^{8}$ V/m

6 0

3 years ago