A marble dropped from a bridge strikes the water in 6.0 seconds

A school bus moves slower and slower. Using what you have learned about forces, explain why the bus moves slower and slower.

Levart [38]

Answer:

I don't exactly know what you learned but it could be because of more friction or the bus was running out of gas.

4 0

3 years ago

If a 70kg swimmer pushes off a pool wall with a force of 250N. What is her acceleration?

Alenkasestr [34]

Newton taught us: Force = (mass) x (acceleration)

Divide each side by (mass) : Acceleration = (force) / (mass) .

The only problem here is: This formula applies when the "Force" is the
only force on the object. When the objects in these school problems are
falling out of airplanes, shot from guns, or being hit by baseball bats, we
routinely ignore the force of air resistance against the object. We're
comfortable with that, maybe because it's become a habit. But now,
we're not so comfortable about ignoring the force of water resistance.

All I can tell you is that if you DO ignore the water resistance, that is,
if the water were not there, her acceleration would be

(250 newtons) / (70 kg) = 3.57 m/s² = about 0.36 g .

But what is it really, in the water ?

If you've spent any substantial amount of time anywhere near competitive
swimmers, then you know that it depends on their position coming off the
wall, what they do with their knees and knuckles, how straight they hold
their body, how deep the texture of their swim-cap is, and how well they've
shaved their legs.

6 0

4 years ago

Read 2 more answers

Need help in the middle one

kondor19780726 [428]

Answer:

Guysi hate math answer this guy plsss ssss

7 0

3 years ago

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas

SVEN [57.7K]

Answer:

the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s

Explanation:

given information

car's mass, m = 1200 kg

$h_{A}$ = 100 m

$v_{A}$ = $v_{A}$

$h_{B}$ = 150 m

$v_{B}$ = 0

according to conservative energy

the distance from point A to B, h = 150 m - 100 m = 50 m

the initial speed $v_{A}$

final speed $v_{B}$ = 0

thus,

$v_{B}$ ² = $v_{A}$ ² - 2 g h

0 = $v_{A}$ ² - 2 g h

$v_{A}$ ² = 2 g h

$v_{A}$ = √2 g h

= √2 (9.8) (50)

= 31.3 m/s

8 0

3 years ago

If the frequency of the 13C signal of TMS is 201.16 MHz, the two 13C signals of acetic acid at 179.0 and 20.0 ppm are separated

Lelu [443]

The difference in frequency of the two signals is $1.33 \times 10^{10} \ kHz$ .

The given parameters;

<em>frequency of the 13 C signal = 201.16 MHz</em>

The energy of the 13 C signal located at 20 ppm is calculated as follows;

$E = hf\\\\E_1 = h \frac{c}{\lambda} \\\\E_1 = \frac{(6.626 \times 10^{-34})\times 3\times 10^8}{20 \times 10^{-6}} \\\\E_1 = 9.94 \times 10^{-21} \ J$

The energy of the 13 C signal located at 179 ppm is calculated as follows;

$E_2 = \frac{hc}{\lambda} \\\\E_2 = \frac{(6.626\times 10^{-34})\times (3\times 10^{8})}{179 \times 10^{-6} } \\\\E_2 = 1.11 \times 10^{-21} \ J$

The difference in frequency of the two signals is calculated as follows;

$E_1- E_2 = hf_1 - hf_2\\\\E_1 - E_2 = h(f_1 - f_2)\\\\f_1 - f_2 = \frac{E_1 - E_2 }{h} \\\\f_1 - f_2 = \frac{(9.94\times 10^{-21}) - (1.11 \times 10^{-21})}{6.626\times 10^{-34}} \\\\f_1 - f_2 = 1.33 \times 10^{13} \ Hz\\\\f_1 - f_2 = 1.33\times 10^{10} \ kHz$

Thus, the difference in frequency of the two signals is $1.33 \times 10^{10} \ kHz$ .

Learn more here:brainly.com/question/14016376

4 0

3 years ago