Answer:
Explanation:
From the given question, the small sphere was provided with an excess charge of +3 C, while the smaller shell was given an excess of -7 C, it should be -7 C and not 7 C.
So, in light of that, to determine the electric charges values & signs on each of them, we have:
on a = +3 C
on b = -7 C
on c = -7 C
on d = +3 C
on e = -7 C
Answer:
If one bulb goes out, other bulbs stays lit.
If there is break in one branch of the circuit, current can still flow through other branches.
Parallel circuits are simple to design and build.
Explanation:
In parallel circuits all the components and resistors are connected to common terminal or common supply.
So here if we disconnect one of the branch of the circuit then it will not affect the other branches of the circuit so it will not affect the current of other branches.
So here it is very easy to build a parallel circuit and if one of the branch of the circuit goes out then it will not affect the other branches.
Answer:
Explanation:
Given that,
Basket ball is drop from height
H=10m
It is dropped on planet mass
And the acceleration due to gravity on Mars is given as
g= 3.7m/s²
Time taken for the ball to reach the ground
Initial velocity of the body is zero
u=0m/s
Using equation of motion: free fall
H = ut + ½gt²
10 = 0•t + ½ × 3.7 ×t²
10 = 0 + 1.85t²
10 = 1.85t²
Then, t² =10/1.85
t² = 5.405
t = √ 5.405
t = 2.325seconds
So the time the ball spend on the air before reaching the ground is 2.325 seconds
Answer:
Explanation:
According to energy conservation which states that the workdone is equal to change in the system
Workdone = change in kinetic energy + (frictional force * distance)
Workdone = ΔK + fd
Workdone = kf-Ki + fd
Workdone = = 1/2(m(v-u)^2) + fd
Given
Mass m = 495kg
final velocity v = 105m/s
initial velocity = 0m/s
Force f= 1400N
distance d = 395m
Substitute
Workdone = 1/2(495(105-0)^2) + 1400(395)
Workdone = 2,728,687.5+553000
Workdone = 3,281,687.5 Joules
Time = 8.2secs
Power output = Workdone/Time
Power output = 3,281,687.5/8.2
Power output = 885,766.768
Power output = 8.858 * 10^5 watts
CORRECT ANSWER:
a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.
STEP-BY-STEP EXPLANATION:
The complete question from book is
According to Figure 9.6, what is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?
a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.
b- Signaling molecules that bind to cell-surface receptors lead to cellular responses restricted to the cytoplasm; signaling molecules that bind to intracellular receptors lead to cellular responses restricted to the nucleus.
c- Cell-surface receptors bind to specific signaling molecules; intracellular receptors bind any signaling molecule.
d- Cell-surface receptors typically bind to signaling molecules that are smaller than those bound by intracellular receptors.
e- None of the other answer options is correct.
