Marks: 1
What is the power of a light that uses 300 joules of energy in 60
seconds?
18,000 W
50 W
5 W
0.2 W

The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

The given parameters;

Mass of the first object, m1 = 1 kg

Mass of the second object, m2 = 5 kg

The final velocity of the objects during the downward motion is calculated as follows;

[tex]v_f = v_0 + gt\\\\v_f = 0 + gt\\\\\v_f = gt[/tex]

The time of motion of the object from the given height is calculated as;

[tex]h = v_0 t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} }[/tex]

The time of motion of each object is independent of mass of the object.

Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

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Answer:

B

Explanation:

Wave length is the height perpendicular verically of a wave

Answer:

b

Explanation:

makes the most sense to me

Answer:

x=v.t

The answer: Distance= Speed x Time

And also

Time = Distance/Speed

Speed= Distance/Time

Answer:

Explanation:

15 km(1000 m / km) / (20 min(60 s/min)) = 12.5 m/s

Answer:

Light behaves differently in space than on Earth.

Explanation:

Because the gravity field is greater near earth than in most of space. Not the areas near stars, black holes, pulsars, and such but in the vast emptyness between the clumpy spots.

Answer:

Explanation:

constant acceleration???

assume it to be so

average speed is (10 + 30) / 2 = 20 m/s

t = d/v = 60/20 = 3 s

Answer:

Explanation:

momentum is conserved. Initial momentum was zero, so final total momentum must also be zero

0.042(650) + 70v = 0

v = -0.39 m/s

|v| = 0.39 m/s

Answer:

the speed of the object has become constant.

Explanation:

At terminal velocity, air resistance equals in magnitude the weight of the falling object. Because the two are oppositely directed forces, the total force on the object is zero, and the speed of the object has become constant.

Answer:

The frictional force producing this deceleration would have a magnitude of [tex]4\; \rm N[/tex].

Explanation:

The velocity of this object changed by [tex]\Delta v = (-10\; \rm m\cdot s^{-1})[/tex] in [tex]\Delta t = 5\; \rm s[/tex]. The acceleration of this object would be:

[tex]\begin{aligned}a &= \frac{\Delta v}{\Delta t} \\ &= \frac{-10\; \rm m\cdot s^{-1}}{5\; \rm s} = -2\; \rm m\cdot s^{-2}\end{aligned}[/tex].

Let [tex]m[/tex] denote the mass of this object. By Newton's Second Law of Motion, the net force on this object would be:

[tex]\begin{aligned}F &= m \, a \\ &= 2\; \rm kg \times (-2\; \rm m\cdot s^{-2}) \\ &= -4\; \rm N\end{aligned}[/tex].

([tex]1\; {\rm kg \cdot m \cdot s^{-2} = 1\; {\rm N}[/tex].)

If the floor is level, friction would be the only unbalanced force on this object. Thus, the magnitude of the frictional force on this object would also be [tex]4\; {\rm N}[/tex], same as the magnitude of the net force on this object.

Answer:

Explanation:

The work increased the potential energy

W = PE = mgh = 40(9.8)(15) = 5880 J(oules)

Answer:

a = Δv/Δt = (0 - 20) / (5 - 2) = -6⅔ m/s²

The mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.

HOW TO CALCULATE MASS:

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Answer:

The one falling from the greatest height will have the greatest speed.

h = 1/2 g t^2        time for ball to fall distance h

h2 / h1 = t2^2 / t1^2       dividing equations

h2 / t2^2 = h1 / t1^2

Let v be the average speed (v2 = h2 / t2)

1 / t2 * v2 = 1 / t1 * v1

v2 / v1 = t2 / t1      the one taking the longest to fall has the greater av. speed

Check:

h4 / h1 = t4^2 / t1^2     or

t4 / t1 = (h4 / h1)^1/2

In this case t4 / t1 = (4 / 1)^1/2 = 2  or twice the average speed

t1 = (2 h / g)^1/2 = .2^1/2 = .447       using g = 10

t4 = (2 h / g)^1/2 = .8^1/2 = .894

v1 = 1 / .447 = 2.24 m/s average speed

v4 = 4 / .894 = 4.47    or twice the average speed

Answer:

[tex]47 \ \frac{m}{s}[/tex]

Explanation:

s = displacement (m)

u = initial velocity [tex](\frac{m}{s})[/tex]

v = final velocity [tex](\frac{m}{s})[/tex]

a = acceleration [tex](\frac{m}{s^{2} })[/tex]

t = time (s)

s = 235

a = -4.7

v = 0

v² = u² + 2as

(0)² = u² + 2(-4.7)(235)

u² - 2209 = 0

u² = 2209

u = 47

Answer:

[tex]\boxed {\boxed {\sf 47 \ m/s}}[/tex]

Explanation:

We are asked to find the initial velocity of the car before it began to accelerate.

We are given the acceleration, distance, and final velocity, so we will use the following kinematic equation:

[tex]{v_f}^2 = {v_i}^2 + 2ad[/tex]

The car's acceleration is -4.7 meters per second square. It traveled a distance of 235 meters. It came to rest, or a final velocity of 0 meters per second.

Substitute the values into the formula.

[tex](0 \ m/s)^2 = {v_i}^2 + 2 (-4.7 \ m/s^2)(235 \ m)[/tex]

[tex]0 = {v_i}^2 + 2 (-4.7 \ m/s^2)(235 \ m)[/tex]

Multiply the numbers in parentheses.

[tex]0= {v_i}^2 + (-2209 \ m^2 / s^2)[/tex]

Add -2209 to both sides of the equation.

[tex]0+ 2209 \ m^2 /s^2 = {v_i}^2+ ( -2209 \ m^2 /s^2 )+ 2209 \ m^2 /s^2[/tex]

[tex]2209 \ m^2 /s^2 = {v_i}^2[/tex]

Take the square root of both sides.

[tex]\sqrt {2209 \ m^2 /s^2} = \sqrt {{v_i}^2[/tex]

[tex]\sqrt {2209 \ m^2 /s^2} = v_i[/tex]

[tex]47 \ m/s = v_i[/tex]

The inital velocity of the car was 47 meters per second.

Hi there!

Converting from angular velocity to tangential velocity can be done by:

v = ωr

v = tangential velocity (m/s)

ω = angular velocity (rad/sec)

r = radius (m)

Convert 12 cm to meters:

100 cm = 1 m

12 cm = 0.12 m

Now, convert rev/min to rad/sec:

[tex]{\frac{200rev}{min}} * \frac{1min}{60s} * \frac{2\pi rad}{1 rev} = 20.94 rad/sec[/tex]

v = 20.94 · 0.12 = 2.51 m/s

Explanation:

12N by first law of newton is net force after colloision

Answer:

(a) The force changes its magnitude with respect to displacement, hence the total work will be sum of increment of work in three steps:-

step 1 . from 0 to 0.25m .

force = 0.6 N

displacement= 0.25m

work done =( force × displacement) = (0.25 × 0.6 ) = 0.15 joule.

step 2:- .

work done in moving from 0.25 to 0.50 m.

work done = ( force × displacement) = ( 0.4 × 0.25) = 0.10 Joule. .

step 3 :-

work done in moving from 0.50 to 0.75 m

work done = 0.8 × 0.25 = 0.200 joule.

hence total work done = ( 0.20+0.10+0.15) = 0.45 joule. ans

(b) similar concept you have to use here also.

step 1:

from 0.20 to 0.50, force of magnitude 0.4 N acts on the object.

Work done = ( 0.50-0.20)× 0.4 = 0.30 × 0.4 = 0.12 joule.

step :- 2

from 0.50 to 0.55 , force of magnitude 0.8 N acts on the block.

work done = 0.8× ( 0.55- 0.50) = 0.04 joule

total work done = 0.04 + 0.12 = 0.16 joule. ans

Answer:

Chronic Condition or Long-term Illness

Explanation:

They can also be known as Chronic Conditions or Long-Term Illnesses, hope this helps.

Answer:

Chronic condition, also called long-term condition.

Explanation:

0.8 seconds

Explanation:

time of flight = 2u/g

u=4m/s

g=10

= 8/10

= 0.8 sec

just a trial...not sure!!!

Given :

∅ = 60⁰

u = 4 m/s

g = 10m/s²

to find :

T = ?

Solution :

as per formula,

[tex]t = \frac{2u \: sin \theta}{g} [/tex]

now put the value : [tex]t \: = \frac{2 \times 4 \times sin \: 60}{10} [/tex]

as we know [tex] sin60 \: = \frac{ \sqrt{3} }{2} [/tex]

therefore,

[tex]t \: = \frac{8 \times \frac{ \sqrt{3} }{2} }{10} [/tex]

as we solve this we get,

[tex]t \: = \frac{ 2\sqrt{3} }{5} [/tex]

that's t = 0.69 sec

[tex]\sf\fbox\red{\:I \:hope \:it's \:helpful \:to \:you}[/tex]

Answer:

Explanation:

The mechanical advantage of a pulley with three ropes is one.

is incorrect. It is 3

Answer:

b is the ans......

Explanation:

"b" (and any subsequent words) was ignored because we limit queries to 32 words.

Answer:

Explanation:

T = 2π[tex]\sqrt{L/g}[/tex]

(T / 2π)² = L/g

g = 4π²L/T²

g = 4π²(0.75000)/(1.7357)²

g = 9.82814766...

g = 9.8281 m/s²

Rewrite the amounts as improper fractions:

9 2/3 = 29/3

3 7/8 = 31/8

Rewrite both fractions with a common denominator

29/3 = 232/24

31/8 = 93/24

Now subtract: 232/24 - 93/24 = 139/24

Rewrite as a proper fraction: 5 19/24

Answer 5 19/24

Answer:

kinetic energy is given as KE = (0.5) m v²given that : v = speed of the bottle in each case =  4 m/s when m = 0.125 kg KE = (0.5) m v² =  (0.5) (0.125) (4)²

Explanation:

Answer:

1. 0.5    2.  2    3. 3.75    4.  5

Explanation:

The locations where the riders feel less than their normal weight are Location A, Location C and Location D.

The given parameters;

The normal weight of the riders is calculated by applying Newton's second law of motion as follows;

W = mg

W = 9.8m

The apparent weight of the riders for the upward acceleration is calculated  as follows;

[tex]R = m(g + a)[/tex]

The apparent weight of the riders for the downward acceleration is calculated  as follows;

[tex]R = m(g - a)[/tex]

The apparent weight of the riders at location A is calculated as follows;

[tex]R_ A = m(9.8 - 10)\\\\R_ A = -0.2 m[/tex]

The apparent weight of the riders at location B is calculated as follows;

[tex]R_B = m(9.8 + 2)\\\\R_B = 11.8 m[/tex]

The apparent weight of the riders at location C is calculated as follows;

[tex]R_C = m(9.8 - 6)\\\\R_C = 3.8 m[/tex]

The apparent weight of the riders at location D is calculated as follows;

[tex]R_D = m(9.8 - 12)\\\\R_D = -2.2 m[/tex]

The apparent weight of the riders at location E is calculated as follows;

[tex]R_E = m(9.8 + 6)\\\\R_E = 15.8 m[/tex]

Thus, the locations where the riders feel less than their normal weight are;

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Answer:

35.2644

I suppose mew is refractive index

Explanation:

( sin i ) / (sin r) = refractive index

( sin 60) / (sin r) = 1.5

( sin 60) / 1.5 =sin r

r=35.844

sorry if I'm wrong

Answer:

An orbit of a spacecraft from which the spacecraft or another vehicle may be launched on a new trajectory.

Answer:

The types of energy is bond breaking and bond forming in chemical energy.

Explanation:

During Chemical reaction energy is required either for breaking up bonds in case of reactants and building bonds to form products.

The chemical reaction in which energy is released is called exothermic reactions, which is released due to making up the bonds.

The chemical reaction in which energy is absorbed is called endothermic reactions, in which energy is absorbed for breaking up the bonds.