Ask question

Ask question

All categories

3 years ago

10

Two object, one of mass m and the other of mass 2m, are dropped from the top of a building. if there is no air resistance, when

they hit the ground

1 answer:

ra1l [238]3 years ago

3 0

Answer:

They hit the ground at the same time

Explanation:

Mass doesn't matter in free fall. As long as they were released with the same velocity and the same height, they'll hit the ground at the same time

You might be interested in

A long, straight wire carries a current of 8.60

Kay [80]

First of all, we need to calculate the magnetic field generated by the wire at a distance r=3.90 cm=0.039 m, which is given by:

$B=\frac{\mu_0 I}{2 \pi r}$

where I=8.60 A is the current. Substituting numbers in the equation, we find

$B=\frac{(4 \pi \cdot 10^{-7})(8.6 A)}{2 \pi (0.039 m)}=4.4 \cdot 10^{-5} T$

And now we can calculate the force exerted on the electron, which is given by:

$F=qvB$

where $q=1.6 \cdot 10^{-19} C$ is the charge of the electron and $v=5.0 \cdot 10^4 m/s$ is its speed. Substituting data in the formula, we find

$F=(1.6 \cdot 10^{-19}C)(5.00 \cdot 10^4 m/s)(4.4 \cdot 10^{-5} T)=3.5 \cdot 10^{-19} N$

6 0

3 years ago

Rain and wind place which type of load on structures?

katovenus [111]

It’s a dynamic load.
Hopes this helps :)

4 0

3 years ago

Read 2 more answers

In 1996, astronomers discovered an icy object beyond pluto that was given the designation 1996 tl 66. it has a semimajor axis of

antoniya [11.8K]

Answer : 2446 years.

Explanation :

Length of semi major axis is, $a=84\ au= 1.496\times 10^{11}\ m$

According to Kepler's third law, square of time period of an orbit is directly proportional to the cube of the semi major axis.

i.e $T^2=\dfrac{4\pi^2}{GM}a^3$

where G is gravitational constant

M is mass of sun, $M=1.98\times 10^{30}\ Kg$

So, $T^2=\dfrac{4\times (3.14)^2}{6.6\times 10^{-11}Nm^2/Kg\times 1.98\times 10^{30}\Kg}$

$T^2=3\times 10^{-19}\times(84\times 1.496\times 10^{11})^3$

$T^2=3\times 10^{-19}\times 1984415.6\times 10^{33}$

$T^2=59532469.8\times 10^{14}\ s$

$T=7715.7\times 10^7\ s$

since, $1\ sec=3.17\times 10^{-8}\ years$

So, orbital period is approximately 2446 years.

7 0

4 years ago

Calculate the mass of 1.35 moles of sodium chloride (NaCl).

Harlamova29_29 [7]

Answer:d

Explanation:

4 0

3 years ago

A tightrope walker wants to know the tension used to support the rope that is suspended between two poles. The rope is 18 m long

Alla [95]

Answer:

T = 29.6 N

Explanation:

length of the rope is

L = 18 m

mass of the rope is

m = 12 kg

now we have

mass per unit length of the rope is given as

[te]\lambda = \frac{12 kg}{18 m}[/tex]

now time taken by wave to reach from end to other

$t = \frac{L}{v}$

$2.7 s = \frac{18}{v}$

$v = 6.67 m/s$

now we have

$v = \sqrt{\frac{T}{\lambda}}$

$6.67 = \sqrt{\frac{T}{0.67}}$

so we will have

$T = 29.6 N$

5 0

4 years ago