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butalik [34]
3 years ago
15

Spaceship 1 and Spaceship 2 have equal masses of 300 kg. Spaceship 1 has

Physics
1 answer:
UkoKoshka [18]3 years ago
4 0

Answer:

30 m/s

Explanation:

Momentum = mass × speed

9000 kg m/s = (300 kg) v

v = 30 m/s

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You have probably noticed that carrying a person in a pool of water is much easier than carrying a person through air. To unders
Kruka [31]

Answer:

The answer is "0.91238 and 744.8"

Explanation:

In this scenario it is easier to take a person to the water-pool than to transport the people in the air, as the person's strength is increased by water upwards:

f_b \to m \to mg =person \\\\F_B \ in\  air = v\ & air\  g \\\\

               =0.076 \times 1.225 \times 9.8 \\\\ =0.91238 \ N\\\\

F_B \ in \ water = v  \& water \ g \\\\

                    =0.076 \times 1000 \times 9.8\\\\= 744.8 \ N\\

6 0
3 years ago
When you stretch a spring 20 cm past its natural length, it exerts a force of 8
Nastasia [14]

Answer:

40 N/m

Explanation:

F = -kx (This is the Hooke's Law equation)

F is the force the spring exerts = 8 N

-k = spring constant

x = displacement (The distance stretched past it's natural length) = 20cm

x needs to be in meters, and 20 cm is = to 0.2 meters

Finally:

8N = -k (0.2m)

-k = 8N / 0.2 m

k = -40 N/m

6 0
3 years ago
Read 2 more answers
A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T.
sineoko [7]

Answer:10842.33m/s

Explanation:

F=qvBsine

V=f/(qBsine)

V=(3.5×10^-2)÷(8.4×10^-4×6.7×10^-3×sin35)

V=10842.33m/s

5 0
3 years ago
if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

Therefore, the mutual force between the two point charges is 319.64 N

4 0
3 years ago
An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m, wh
lara [203]

Answer:

a_y = 4.9\ m/s^2

Explanation:

Given,

Width of rectangular tank, b = 1 m

Length of the tank, l = 2 m

height of the tank, d = 1.5 m

Depth of gasoline on the tank, h = 1 m

\dfrac{dz}{dy}=-\dfrac{1.5-1}{1}

\dfrac{dz}{dy}=-0.5

The differential form with the acceleration

\dfrac{dz}{dy}=\dfrac{-a_y}{a_z + g}

-0.5=-\dfrac{a_y}{a_z + g}

acceleration in z-direction = 0 m/s²

g = 9.8 m/s²

a_y is the horizontal acceleration of the gasoline.

0.5=\dfrac{a_y}{0 + 9.8}

a_y = 9.8\times 0.5

a_y = 4.9\ m/s^2

Hence, Horizontal acceleration of the gasoline before gasoline would spill is equal to 4.9 m/s²

3 0
3 years ago
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