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3 years ago

Given the picture below, which bullet was fired first?

Physics

2 answers:

mart [117]3 years ago

7 0

B was fired first and just trust me

olchik [2.2K]3 years ago

5 0

Answer:

Explanation:

i think

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Question 7 of 10

astra-53 [7]

Answer D is an example of melting

8 0

3 years ago

Read 2 more answers

The picture below shows an asteroid and a comet.

Tom [10]

Answer:

01) Asteroids are formed closer to the sun than comets.

Explanation:

Majority of asteroids are present region between Jupiter and Mars while the comets are found away from the orbits of Pluto and Neptune.

6 0

4 years ago

The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -41 oC and 900 Pa, respect

Fittoniya [83]

Answer:

Part a)

$\rho = 0.0205 kg/m^3$

Part b)

$\rho = 1.22 kg/m^3$

So density of atmosphere at Martian Surface is very less than the density at Earth.

Explanation:

Part a)

As per ideal gas equation we know that

$PM = \rho RT$

here we know that Martian atmosphere is equivalent to that of carbon

so we will have

$P = 900 Pa$

$T = 273 - 41 = 232 K$

now we will have

$(900)(0.044) = \rho (8.31)(232)$

$\rho = 0.0205 kg/m^3$

Part b)

Now for the earth surface the density of air is given for

$P = 101.6 kPa$

$T = 18 ^oC$

so we will have

$PM = \rho RT$

$(101.6\times 10^3)(0.029) = \rho(8.31)(273 + 18)$

$\rho = 1.22 kg/m^3$

So density of atmosphere at Martian Surface is very less than the density at Earth.

8 0

3 years ago

URGENTE. ¿Qué fuerzas actúan en un tobogán?

professor190 [17]

Answer:

friction or fricción

if u only speak spanish

Explanation:

5 0

3 years ago

A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se

Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

$v = \frac{Bqr}{m}$

$v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}$

v = 661098.9 = 661099 m/s

(b)

$\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}$

e / m = 1.76 x 10^14 C / kg

K = 0.5 x mv²

K = 0.5 x 9.1 x 10^-31 x 661099 x 661099

K = 1.99 x 10^-19 J

K = 1.24 eV

So, the potential difference is

V = 1.24 V

(d) if the acceleration voltage is doubled

V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J

Let v is the speed

K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

$r=\frac{mv}{Bq}$

$r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}$

r = 0.0113 m = 1.13 cm

7 0

3 years ago