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3 years ago

Your bike is too big if your can't touch the ground A.Knees B.Hands C.Feet

Physics

1 answer:

SOVA2 [1]3 years ago

6 0

Answer: C: Feet

Explanation: Why...

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Suppose a parachutist is falling toward the ground, and the downward force of gravity is exactly equal to the upward force of ai

Serhud [2]

Answer:

Option b. is correct.

Explanation:

In the given question a parachutist is falling toward the ground .

Also, the downward force of gravity is exactly equal to the upward force of air resistance.

So, net force applied to the parachutist is equal to zero ( because both force acts in opposite direction ).

Now by first law of motion :

An object will be in rest or in constant speed unless and until no external force is applied on it .

So, in the question the velocity of the parachutist is not changing with time.

Therefore, option b. is correct.

Hence, this is the required solution.

7 0

3 years ago

Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI)

Ludmilka [50]

Answer:

a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.

Explanation:

a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:

$a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$a' = 3.569$

The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:

$v_{f} = v_{o} + a \cdot t$

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v_{f}$ - Final speed, measured in meter per second.

$a$ - Acceleration, measured in $\frac{m}{s^{2}}$ .

$t$ - Time, measured in seconds.

$t = \frac{v_{f}-v_{o}}{a}$

$t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)}{35\,\frac{m}{s^{2}} }$

$t = 0.095\,s$

Lastly, the severity index is now determined:

$SI = \frac{a'^{5}}{2\cdot t}$

$SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}$

$SI = 3047.749$

b) The initial and final speed of the injured are $1.944\,\frac{m}{s}$ and $5.278\,\frac{m}{s}$ , respectively. The travelled distance can be determined from this equation of motion:

$v_{f}^{2} = v_{o}^{2} + 2\cdot a \cdot \Delta s$

Where $\Delta s$ is the travelled distance, measured in meters.

$\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}$

$\Delta s = \frac{\left(5.278\,\frac{m}{s} \right)^{2}-\left(1.944\,\frac{m}{s} \right)^{2}}{2\cdot \left(35\,\frac{m}{s^{2}} \right)}$

$\Delta s = 0.345\,m$ .

8 0

3 years ago

An object weighs 15N in air and 13N when completely

anastassius [24]

Answer:

The volume of water displaced = 0.204 m³

Explanation:

From Archimedes' principle,

Upthrust = lost in weight = weight of water displaced.

U = W₁ - W₂..................... Equation 1

Where U = upthrust, W₁ = weight in air, W₂ = weight when completely submerged in water.

Given: W₁ = 15 N, W₂ = 13 N

Substituting these values into equation 1

U = 15 - 13

U = 2 N

But,

Weight of water displaced = mass of water displaced. × acceleration due to gravity.

Mass of water displaced = weight of water displaced/ acceleration due to gravity.................... Equation 2

Where: acceleration due to gravity = 9.8 m/s², weight of water displaced = 2 N

Substituting these values into equation 2

Mass of water displaced = 2/9.8

Mass of water displaced = 0.204 kg.

Also,

Volume of water displaced = mass of water displaced/ Density of water.............. Equation 3

Where Density of water = 1.0 kg/m³, mass of water displaced = 0.204 kg

Substituting these values into equation 3 above,

Volume of water displaced = 0.204/1

volume of water displaced = 0.204 m³

Therefore the volume of water displaced = 0.204 m³

8 0

3 years ago

the velocity of a car traveling in the positive direction decreases from 32 m/s to 24 m/s in 4 seconds. what is the average acce

anastassius [24]

Answer:

$a=-2m/sec^2$

Negative sign shows that velocity of the car is decreases at a constant rate

Explanation:

We have given velocity of the car is decreases from 32 m /sec to 24 m/sec in 4 second

So initial velocity of the car u = 32 m /sec

And finally car reaches to a velocity of 24 m/sec

Time taken to change in velocity = 4 sec

So final velocity v = 24 m/sec

From first equation of motion v = u+at

So $24=32+a\times 4$

$a=-2m/sec^2$

Negative sign shows that velocity of the car is decreases at a constant rate

6 0

3 years ago

Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative

SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

Explanation:

Given there are three blocks of masses $M_{a}$ , $M_{b}$ and $M_{c}$ (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *a ................(equation 1)

Also it is given that $M_{a}$ does not move with respect to $M_{c}$ , which gives tension T is exerted on pulley by $M_{a}$ only, Hence tension T is

T = $M_{a}$ *a ..........(equation 2)

There is also also tension exerted by $M_{b}$ . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = $M_{b}$ $\sqrt{a^{2} +g^{2} }$ ................(equation 3)