All categories

3 years ago

Jane walked 5.00 meters on a road that inclines 13.0 degrees. How much distance did she cover horizontally?

Physics

1 answer:

Paladinen [302]3 years ago

8 0

Any options available?

You might be interested in

Urgent! please help!

kap26 [50]

$_{94}^{239}Pu+ _{0}^{1}n--->_{40}^{100}Zr+_{x}^{y}Element+2_{0}^{1}n\\ \\ 94+0=40+x+0, x= 54,\\ \\ 239+1 = 100+y+2*1, y=138\\ \\_{54}^{138}Element= _{54}^{138}Xe$

Answer is A.

8 0

3 years ago

Two infinitely long parallel wires carry current in opposite directions. Wire 1 has current 15.0 A and wire 2 has current 19.9 A

natita [175]

Answer:

x= 2*I1*d/(I1+I2) meter

4 0

3 years ago

The Earth, the Sun, and the rest of the solar system are almost unimaginably old when viewed on a human time scale. While modern

Zielflug [23.3K]

While modern humans first evolved approximately 200,000 years ago,
the age of the Earth, the Sun, and the rest of the solar system is believed
to be approximately 4.6 billion years.

That makes the Earth, the Sun, and the rest of the solar system
something like 23 thousand times as old as the human species !

8 0

3 years ago

Read 2 more answers

How does a physicist answer a scientific question?

egoroff_w [7]

Answer:

He does an experment.

6 0

3 years ago

A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary

Ann [662]

The distance traveled by the wood after the bullet emerges is 0.16 m.

The given parameters;

mass of the bullet, m = 23 g = 0.023 g
speed of the bullet, u = 230 m/s
mass of the wood, m = 2 kg
final speed of the bullet, v = 170 m/s
coefficient of friction, μ = 0.15

The final velocity of the wood after the bullet hits is calculated as follows;

$m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s$

The acceleration of the wood is calculated as follows;

$\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2$

The distance traveled by the wood after the bullet emerges is calculated as follows;

$v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m$

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.

Learn more here:brainly.com/question/15244782

7 0

3 years ago