Before beginning an exercise program a person should always

Harry and ron set up this experiment with a glider, whose mass they have measured to be 565 g, and seven washers hanging from th

svetlana [45]

Let's call $m=565~g=0.565~kg$ the mass of the glider and $m_w=7\cdot12~g =84~g=0.084~kg$ the total mass of the seven washers hanging from the string.
The net force on the system is given by the weight of the hanging washers:
$F_{net} = m_w g$
For Newton's second law, this net force is equal to the product between the total mass of the system (which is $m+m_w$ ) and the acceleration a:
$F_{net}=(m+m_w)a$
So, if we equalize the two equations, we get
$m_w g = (m+m_w)a$
and from this we can find the acceleration:
$a= \frac{m_w g}{(m+m_w)} = \frac{0.084~kg \cdot 9.81~m/s^2}{(0.565~kg+0.084~kg)}=1.27~m/s^2$

5 0

3 years ago

I need to figure out A but i’m not sure how

Damm [24]

Answer:

cc

Explanation:

4 0

3 years ago

A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista

slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

$C' = \frac{1}{2}C$

The potential difference across the capacitor is given by

$V= \frac{Q}{C}$

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

$V' = 2 V$

Now we can analyze the electric field between the plates of the capacitor, which is given by

$E=\frac{V}{d}$

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

$E'=\frac{2V}{2d}=\frac{V}{d}=E$

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

4 0

3 years ago

To minimize signal distortion, at each end of the J-1939 CAN-bus there is a(n)_____________resistor.

EastWind [94]

5-ohm
Extra
Variable
120-ohm
Variable
Pg. 614

8 0

3 years ago

A wheel has a constant angular acceleration of 4.5 rad/s2. during a certain 5.0 s interval, it turns through an angle of 128 rad

dalvyx [7]

The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s

3 0

4 years ago