All categories

3 years ago

A 10 kg boat is moving 3 m/s. Find kinetic energy.

Physics

2 answers:

mr Goodwill [35]3 years ago

7 0

KE = 1/2mv^2
m=10 v=3
KE=1/2*10*3^2
KE= 45 Joules

REY [17]3 years ago

4 0

answer

45J

explanation

kinetic energy can be found through the equation $K = \frac{mv^2}{2}$

plug in 10 for m (mass) and 3 for v (velocity)

$K = \frac{mv^2}{2}\\K = \frac{10*3^2}{2}\\K = \frac{10*9}{2}\\K = \frac{90}{2}\\K = 45 J$

You might be interested in

Define unit aland how many types of unit are there . Name them?

Lana71 [14]

Answer:

7. They arethe meter (m), the kilogram (kg), the second (s), the kelvin (K), the ampere (A), the mole (mol), and the candela (cd)

Explanation:

7. They arethe meter (m), the kilogram (kg), the second (s), the kelvin (K), the ampere (A), the mole (mol), and the candela (cd)

4 0

2 years ago

Temperature Dependence of the pH of pure Water

Mamont248 [21]

Answer:

At 100°C, the pH of pure water is 6.14. That is the neutral point on the pH scale at this higher temperature. A solution with a pH of 7 at this temperature is slightly alkaline because its pH is a bit higher than the neutral value of 6.14.

Explanation:

nonde

8 0

3 years ago

10.

nlexa [21]

Answer:

Cell Membrane

Explanation:

The cell membrane controls what goes in and out of a cell, and keeps it shape, much like a city limit.

3 0

3 years ago

You decide to impress Grandpa by showing him how fast sound travels. You have a piece of plastic pipe with an adjustable closed

fredd [130]

Answer:

336.96m/s

Explanation:

answer is in photo above

3 0

3 years ago

Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=

drek231 [11]

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

$v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2$

$v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2$

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

$v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

$v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

The final relative velocity of the satellite, $v_f$ = v₁ + v₂

∴ $v_f$ = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, $v_f$ = 0.190 m/s

4 0

2 years ago