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Firdavs [7]
3 years ago
13

Rutherford proposed that

Physics
1 answer:
Mila [183]3 years ago
4 0

The answer is C. Because Rutherford purposed that negatively charged electrons orbit a positively charged nucleus in orbits with set energy levels.

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The acceleration due to gravity is 9.81 m/s2, towards the Earth. Rain falling from an altitude of 9,000 m would fall for about 1
Anna [14]

Answer:

the final speed of the rain is 541 m/s.

Explanation:

Given;

acceleration due to gravity, g = 9.81 m/s²

height of fall of the rain, h = 9,000 m

time of the rain fall, t = 1.5 minutes = 90 s

Determine the initial velocity of the rain, as follows;

h = ut + \frac{1}{2} gt^2\\\\9000 = 90u +  \frac{1}{2} (9.8)(90)^2\\\\9000 = 90u + 39690\\\\90u = -30690\\\\u = \frac{-30690}{90} \\\\u = -341 \ m/s

The final speed of the rain is calculated as;

v^2 = u^2 + 2gh\\\\v^2 = (-341)^2 + 2(9.8\times 9000)\\\\v^2 = 292681\\\\v = \sqrt{292681} \\\\v = 541 \ m/s

Therefore, the final speed of the rain is 541 m/s.

3 0
3 years ago
Read 2 more answers
An us bomber is flying horizontally at 300 mph at an altitude of 610 m. its target is an iraqi oil tanker crusing 25kph in the s
Pachacha [2.7K]

Answer:

=1419.19 meters.

Explanation:

The time it takes for the shell to drop to the tanker from the height, H =1/2gt²

610m=1/2×9.8×t²

t²=(610m×2)/9.8m/s²

t²=124.49s²

t=11.16 s

Therefore, it takes 11.16 seconds for a free fall from a height of 610m

Range= Initial velocity×time taken to hit the tanker.

R=v₁t

Lets change 300 mph to kph.

=300×1.60934 =482.802 kph

Relative velocity=482.802 kph-25 kph

=457.802 kph

Lets change 11.16 seconds to hours.

=11.16/(3600)

=0.0031 hours.

R=v₁t

=457.802 kph × 0.0031 hours.

=1.41918 km

=1.41919 km × 1000m/km

=1419.19 meters.

3 0
3 years ago
PROBLEM 5 (Problem 4-145 in 7th edition) Consider a well-insulated horizontal rigid cylinder that is divided into two compartmen
kari74 [83]

Answer:

The answer is "\bold{83.8^{\circ} \ C}".

Explanation:

Formula for calculating the mass in He:

\to m = \frac{PV}{RT}\\

        = \frac{500 \times 1}{ 2.0769 \times (40 + 273)}\\\\ = \frac{500 }{ 2.0769 \times 313}\\\\ = \frac{500 }{ 650.0697}\\\\= 0.76914 \ Kg

Formula for calculating the mass in N_2:

\to m = \frac{PV}{RT}\\

        = \frac{500 \times 1}{ 0.2968 \times (120+ 273)}\\\\ = \frac{500 }{ 0.2968 \times 393}\\\\ = \frac{500 }{ 116.6424}\\\\= 4.2866\ Kg

by using the temperature balancing the equation:

T' = \frac{mcT (He) + mcT ( N_2 )}{ mc (He) + mc ( N_2)}

    = \frac{0.76914 \times 3.1156 \times 313 + 4.2866 \times 0.743 \times393}{  0.76914 \times 3.1156 + 4.2866 \times 0.743} \\\\ = 357  \ \ K \approx  83.8^{\circ} \ C

3 0
3 years ago
Jeremy walked at an average rate of speed of 5 km/h. how far had Jeremy walked in 15 minutes?
astraxan [27]

Answer:

1 half kilometer is right answer

5 0
3 years ago
Read 2 more answers
The position of a 50 g oscillating mass is given by x(t)=(2.0cm)cos(10t−π/4), where t is in s. If necessary, round your answers
Leona [35]
For the answer to the question above, this is the maximum displacement, the spring has only elastic potential energy.
spring is constant @ 5 N/m
maximum displacement = 2 cm = 0.02 m
elastic potential energy = 1/2 kx²
= 0.5 x 5 x 0.02²
So the answer would be
= 0.001 Joules
5 0
3 years ago
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