What is the approximate average speed of the players run to first base

According to Ohm's law, a circuit with a high resistance _____. will have a low electric current will have a high electric curre

lyudmila [28]

Remark
When you are asked a question like this, the first thing to do is search out a formula and put some limits on it.

Formula
I = E/R which comes from E = IR. To get to the derived formula, divide both sides by R
E/R = I*R/R
E/R = I

Discussion
This is an inverse relationship. That means that as one goes up the other one will go down.

So in this case you keep E constant and you manipulate R and look at your results for I

Case 1
Let us say that E = 10 volts
Let us also say the R = 10 ohms

I = E/R
I = 10/10
I = 1 ohm

Case Two
Let's raise the Resistance to 100 ohms
E = 10
R = 100
I = 10/100 = 0.1

Conclusion
As the Resistance goes up, the current goes down. Answer: A

8 0

3 years ago

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Equation of motion description is v=20+2t.How big is the initial speed,acceleration?

trasher [3.6K]

Answer:

the initial velocity is 20 m/s and the acceleration is 2 m/s²

Explanation:

Given equation of motion, v = 20 + 2t

If V represents the final velocity of the object, then the initial velocity and acceleration of the object is calculated as follows;

From first kinematic equation;

v = u + at

where;

v is the final velocity

u is the initial velocity

a is the acceleration

t is time of motion

If we compare (v = u + at) to (v = 20 + 2t)

then, u = 20 and

a = 2

Therefore, the initial velocity is 20 m/s and the acceleration is 2 m/s²

4 0

3 years ago

Please help me out !!!!!!

nikdorinn [45]

Answer:

89

Explanation:

8 0

3 years ago

a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its

lilavasa [31]

Answer:

D/H =15

Explanation:

We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
At this point, we can find the value of H, applying the following kinematic equation:

$v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)$

If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

$H = \frac{v_{0} ^{2} }{2*g} (2)$

We can use the same equation, to find the value of D, as follows:

$v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)$

In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
When we replace these values in (1), we find that v₁ = -v₀.
Replacing in (3), we have:

$(4*v_{0})^{2} - (-v_{0}) ^{2} = 2* g* D\\ \\ 15*v_{0}^{2} = 2*g*D$

Solving for D:

$D = \frac{15*v_{0} ^{2} }{2*g}$

From (2) we know that H can be expressed as follows:

$H = \frac{v_{0} ^{2} }{2*g}$

⇒ D = 15 * H

$\frac{D}{H} = 15$

3 0

3 years ago

Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10

gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

$\Sigma \tau=0$

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

$-\tau_l+\tau_p+\tau_{cm}=0$ (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

$\tau=Fd$ (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

$-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N$

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

$F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N$

The force applied by the left support is 2123.33 N

8 0

3 years ago