In normal light conditions, how well do plants grow when there are 10 plants in the

Giving 20 points and brainiest But please help me

monitta

Answer:

yeah, Do you want me to check your answers? Yes is correct

8 0

3 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

4 years ago

A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal

PilotLPTM [1.2K]

Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
X: mgsin65 - F = mAx 
Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) 
Moment about center of mass: 
Fr = Iα 
Now Ax = rα 
and F = umgcos65 
mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) 
umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) 
ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) 
ugcos65 = 0.4(gsin65 - ugcos65) 
1.4ugcos65 = 0.4gsin65 
1.4ucos65 = 0.4sin65 
u = 0.4sin65/1.4cos65 
u = 0.613

3 0

3 years ago

Find the magnitude and direction of a force between a 25.0 coulomb charge and a 40.0coulomb charge when they are separated by a

deff fn [24]

Answer:

95.0 colomb

Explanation:

Make sure to understand the concept

6 0

3 years ago

A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te

Nataly [62]

Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

$\Delta L=\alpha L_{0}\Delta T$

Where:

L(0) is the initial length
ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
ΔL is the final length minus the initial length (L(f)-L(0))
α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)

So, we have:

$L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)$

$L_{f}=0.117\: m$

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

7 0

3 years ago