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Vedmedyk [2.9K]
3 years ago
15

6. A man is riding a bike with an acceleration of 5.0 meters per second and a mass of the bike and him combined is 20 kg. What f

orce is produced?
Physics
1 answer:
KonstantinChe [14]3 years ago
3 0

Newton's 2nd law of motion:

                         Force = (mass) x (acceleration)

For this man
on his bike:        Force = (20 kg) x (5 m/s²)

                                    =    100 newtons .

This is NOT the "force produced".

This is the force NEEDED in order to accelerate THAT man
and his bike at THAT rate.  If you don't have something that
can apply THAT force, then you can't create this situation,
and he'll just have to carry on, in the same direction and at
the same speed, without turning, slowing down, or speeding up.

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15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and di
lozanna [386]

Answer:

we see it is a linear relationship.

Explanation:

The magnetic flux is u solenoid is

      B = μ₀ N/L   I

where N is the number of loops, L the length and I the current

By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops

      B = (μ₀ I / L)  N

the amount between paracentesis constant, in the case of 4 loop the field is worth

      B = cte 4

N       B

4       4 cte

3       3 cte

2       2 cte

1        1 cte

as we see it is a linear relationship.

In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,

6 0
4 years ago
Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiment
Drupady [299]

Answer:

u_K=0.862

Explanation:

The force of friction between the quails feet and the ground is:

F=m*a

F_K=m*a

F_K=u_k*m*g

u_K*m*g=m*a_c

u_K*g=a

u_K=\frac{a_c}{g}

a_c=\frac{v^2}{r}

So the coefficient of static is solve

u_K=\frac{\frac{v^2}{r}}{g}

u_K=\frac{v^2}{r*g}=\frac{(2.6m/s)^2}{0.80m*9.8m/s^2}

u_K=0.862

4 0
3 years ago
Identify the kind of simple machine represented by each of these examples: a. A drill bit b. A skateboard ramp c. A boat oar
marusya05 [52]

Answer:

a skateboard ramp

Explanation:

6 0
3 years ago
vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses
dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}

(This is correct because the horizontal motion has acceleration zero). Then:

v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}

Then, plugging in the given values, we obtain:

k=\frac{(61.2kg)(15.0m/s)^2}{(2.76m)^2}\\\\k=1808N/m

Finally, the effective spring constant of the firing mechanism is 1808N/m.

3 0
3 years ago
two identical-looking, large, round balls are placed in front of you. one is filled with feathers and the other is filled with s
Lelu [443]
If you apply a little bit of force, one will move easier than the other since it is lighter.
3 0
3 years ago
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