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tester [92]
3 years ago
9

Select the best answer for the question. 6. The first step in using time more efficiently is A. studying how successful people u

se time. B. getting to bed earlier. C. developing a plan. D. limiting recreational activities.
Physics
2 answers:
SIZIF [17.4K]3 years ago
6 0
I would say C.
Developing a plan is important because it will motivate you and tell you when to get things done. It is the most efficient

Hope this helped :)
xxTIMURxx [149]3 years ago
3 0
Hey there!

The first step in using time more effectiantly is to develop a plan. First, you would want to plan how you are going to save your time before you just get a bunch of ideas.

Hope this helps
Have a great day (:
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In the parts that follow select whether the number presented in statement A is greater than, less than, or equal to the number p
Brums [2.3K]

Answer:

a) Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

Explanation:

a) Statement A : 2.567km to two significant figures.

2.567km 2. S.F = 2.6km

Statement B : 2.567km to three significant figures.

2.567km 3 S.F = 2.57km

Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) statement A: (2.567 km + 3.146km) to 2 S.F

(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km

Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).

2.567km to 2 S.F = 2.6km

3.146km to 2 S.F = 3.1km

2.6km + 3.1km = 5.7km

Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

5 0
2 years ago
Examples of increase in pressure due to increase in applied force<br>​
Nina [5.8K]

Answer:

injecting

Explanation:

5 0
3 years ago
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
hammer [34]
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N
therefore, the work done by the weight is
W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J

8 0
2 years ago
An electric heater rated 500 W operates 24 hours/ day. what is the cost of energy to operate it for 20 days at Rs. 2.00 per kWh
Rina8888 [55]

Answer:

Rs. 480.00

Explanation:

1kW = 1000W

therefore 500W = 0.5kW

20 × 24hrs = 480hrs in total.

0.5kW × 480hrs = 240kWh

if rs. 2 for 1kWh

then, 240kWh × 2 = Rs. 480.

4 0
1 year ago
Read 2 more answers
Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0 incline. The coeffi cient of kinetic friction between the bl
UkoKoshka [18]

Answer:

a) W_g =61.25J

b) W_k = -46.25J

c) W_N = 0

d) W_g would be the same.

   W_k would decrease.

   W_N would be the same.

Explanation:

a) On an inclined plane the force of gravity is the sine component of the weight of the block.

F_g = mg\sin(\theta) = 5(9.8)\sin(30^\circ)\\W_g = F_g x = 5(9.8)\sin(30^\circ)2.5 = 61.25J

b) The friction force is equal to the normal force times coefficient of friction.

F_k = -mg\cos(\theta)\mu_k = -5(9.8)cos(30^\circ)0.436 = -18.5 N\\W_k = -F_kx = -46.25J

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.

d) The relation between the vertical height and the distance on the ramp is

h = x\sin(\theta)

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of x\sin(\theta).

The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.

The work done by the normal force would still be zero.

5 0
3 years ago
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