How much work is done by the gravitational force on the block?

Answer:

B, which is why ice skaters often keep their arms close to their body when doing spins and jumps to minimize resistance.

Answer:

480.2 m

Explanation:

The following data were obtained from the question:

Speed of sound (v) = 343 m/s.

Time (t) = 2.8 s

Distance (x) of the cliff =?

The distance of the cliff from the woman can be obtained as follow:

v = 2x /t

343 = 2x /2.8

Cross multiply

2x = 343 × 2.8

2x = 960.4

Divide both side by the coefficient of x i.e 2

x = 960.4/2

x = 480.2 m

Therefore, the cliff is 480.2 m away from the woman.

The distance should be 480.2 m

Since A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. And, there is the velocity of 343 m/s

[tex]v = 2x \div t\\\\343 = 2x \div 2.8\\\\2x = 343 \times 2.8[/tex]

2x = 960.4

x = 480.2 m

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Answer:

the velocity of the acorn

Explanation:

just do in in real life and see

Answer:

it is probably the velocity of the acorn

Answer: 117.6N

Explanation:

By the second Newton's law, we know that:

F = m*a

F = force

m = mass

a = acceleration

We know that in the surface of the Earth, the gravitational acceleration is g = 9.8m/s^2.

Then we just can input that acceleration in the above equation, and also replace m by 12kg, and find that the force due the gravity is:

F = 12kg*9.8m/s^2 = 117.6N

Answer:

v = 0.823 m/s

Explanation:

A man walks south at a speed of 2.00 m/s for 60.0 minutes.

The distance covered in South = 60 × 60 × 2 = 7200 m

He then turns around and walks north a distance 3000 m in 25.0 minutes.

As they moved in opposite direction, net displacement will be : 7200 - 3000 = 4200 m

Average velocity of the man = net displacement/time

[tex]v=\dfrac{4200\ m}{(60+25)\times 60}\\\\=0.823\ m/s[/tex]

So, the average velocity of the man is 0.823 m/s.

Answer:

36.87°

Explanation:

Given

Maximum height = 60m

Horizontal distance (range) = 80,m

Required

Initial speed U

Angle of launch

To get the speed, we will use the range formula;

R = U √2H/g

80 = U√2(60)/9.8

80 = U√12.25

80 = 3.5U

U = 80/3.5

U = 22.86m/s

Get the angle of launch

Using the formula

Theta = tan^-1(y/x)

y is the vertical distance

x is the horizontal distance

Theta = tan^-1(60/80)

Theta = tan^-1(0.75)

Theta = 36.87°

Hence the angle of launch is 36.87°

Answer:

For any given projector, the width of the image (W) relative to the throw distance (D) is know as the throw ratio D/W or distance over width. So for example, the most common projector throw ratio is 2.0. This means that for each foot of image width, the projector needs to be 2 feet away or D/W = 2/1 = 2.0.

[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]

➡ 150hrs.

[tex]{\underline{\pink{\textsf{\textbf{Explanation : }}}}}[/tex]

➡ Time = distance × speed

➡ Time = 15*10

➡ Time = 150hrs ans.

Answer:

Option 1

Explanation:

I always see animals do that

Answer:

476.35 km

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 11000 km/hr

Final velocity (v) = 0 km/hr (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) = ?

Next, we shall convert 9.8 m/s² to km/hr². This is illustrated below:

1 m/s² = 12960 km/hr²

Therefore,

9.8 m/s² = 9.8 m/s² × 12960 km/hr² / 1 m/s²

9.8 m/s² = 127008 km/hr²

Thus, 9.8 m/s² is equivalent to 127008 km/h²

Finally, we shall determine the maximum height reached by the projectile.

This is illustrated below:

Initial velocity (u) = 11000 km/hr

Final velocity (v) = 0 km/hr (at maximum height)

Acceleration due to gravity (g) = 127008 km/hr²

Maximum height (h) = ?

v² = u² – 2gh (since the projectile is going against gravity)

0² = 11000² – (2 × 127008 × h)

0 = 121×10⁶ – 254016h

Collect like terms

0 – 121×10⁶ = – 254016h

– 121×10⁶ = – 254016h

Divide both side by – 254016

h = – 121×10⁶ / – 254016

h = 476.35 km

Thus, the maximum height reached by the projectile is 476.35 km

Answer:

Answer is (b) Mercury, venus and Mars.

Explanation:

i think b is correct!!

;-) :-) :-) :-)

Answer:

option 3

Explanation:

can i get brainliest

Answer:

Explanation:

m1v1=m2v2

m1=70 kg

m2=10 g=0.01 kg

v2=500 m/s

m1v1=m2v2

v1=m2v2/m1

v1=0.01*500/70

v1=0.07

Answer:

220v

Explanation:

Sorry, the question is incomplete

Answer:

on the potential difference applied and on the resistance of the wire.

Explanation:

Ohms law state that the current through a conductor between two points is directly proportional to the potential difference across the two points. Imtroducing the comstant of proportionality, the resistance, one arrives at the usual athematical equation that describes this relationship: I = V/R.

Answer:

Explanation:

Step one:

given data

we are told that 1 large pizza can be cut into 8 even slices

Required

we want to find how many slices are there in 4 large pizzas

Step two:

so if 1 pizza has 8 slices

       4 pizza will have x

cross multiply we have

x= 8*4

x=32 slices

Answer:

B: lodestone

Explanation:

Each magnet has its magnetic poles, north (N) and south (S). Diversified ones are attracted and reptiles of the same name are repelled, similarly to charges, so it was considered possible to separate the magnet at the north and south poles.

Magnetic properties can be lost if the magnet is exposed to high temperatures if it falls or due to some mechanical shocks.

Answer:

Evaporation from oceans = 460,000 km³

Explanation:

Given:

Precipitation on oceans = 420,000 km³

Precipitation on land = 130,000 km³

Evaporation from land = 90,000 km³

Find:

Evaporation from oceans

Computation:

Evaporation from oceans = Precipitation on oceans + Precipitation on land - Evaporation from land

Evaporation from oceans = 420,000 km³ + 130,000 km³ - 90,000 km³

Evaporation from oceans = 460,000 km³

Answer:ccccccC

Explanation:okay

Answer:

2

Explanation:

there are 2 significant figures in there

Answer:

A vibration is a repeated back-and-forth or up-and-down motion.

Explanation:

Waves carry energy through empty space or through a medium without transporting matter.

Answer:

distance i think

Explanation:

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, to determine the bicyclist that traveled the fastest at the end of the race, the speed of the bicyclists at the end of the race will determine this (not the bicyclist that came first nor there overall speed). The speed of the bicyclist at the end of the race can be determined by using the formula below

s = d ÷ t

Where s is the speed of each bicyclist at the end of the race

d is the specific distance covered by the bicyclist at the end of the race

t is the time taken for the bicyclist to complete that distance

It should be noted that to get an accurate result, the distance covered at the end of the race must be the same for all the bicyclists.

Answer:

a) The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty in its perimeter is 0.8 centimeters.

Explanation:

a) From Geometry we remember that the perimeter of the rectangle ([tex]p[/tex]), measured in centimeters, is represented by the following formula:

[tex]p = 2\cdot (w+l)[/tex] (1)

Where:

[tex]w[/tex] - Width, measured in centimeters.

[tex]l[/tex] - Length, measured in centimeters.

If we know that [tex]w = 6.4\,cm[/tex] and [tex]l = 8.3\,cm[/tex], then the perimeter of the rectangle is:

[tex]p = 2\cdot (6.4\,cm+8.3\,cm)[/tex]

[tex]p = 29.4\,cm[/tex]

The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty of the perimeter ([tex]\Delta p[/tex]), measured in centimeters, is estimated by differences. That is:

[tex]\Delta p = 2\cdot (\Delta w + \Delta l)[/tex]  (2)

Where:

[tex]\Delta w[/tex] - Uncertainty in width, measured in centimeters.

[tex]\Delta l[/tex] - Uncertainty in length, measured in centimeters.

If we know that [tex]\Delta w = 0.2\,cm[/tex] and [tex]\Delta l = 0.2\,cm[/tex], then the uncertainty in perimeter is:

[tex]\Delta p = 2\cdot (0.2\,cm+0.2\,cm)[/tex]

[tex]\Delta p = 0.8\,cm[/tex]

The uncertainty in its perimeter is 0.8 centimeters.

Answer:

p₂ / p₁ = 2 (v₁ / v₂)

Explanation:

The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,

         p = m v / √[1+ (v/c)² ]

for the case of speeds much lower than the speed of light this expression is close to

         p = m v

In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small

        p₂ = 2 p₁

       p₂/p₁ = 2

in consecuense

       m v₂ = 2 m v₁

       v₂ = 2 v₁

consider particles of equal mass.

By the time their speeds increase they enter the relativistic regime

        p₂ = mv₂ /√(1 + v₂² /c²)

        p₁ = m v₁ /√(1 + v₁² / c²)

let's look for the relationship between these two moments

       p₂ / p₁ = mv₂ / mv₁   [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)

from the initial statement

      p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)

we take c from the root

      p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]

this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1

      p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]

      p₂ / p₁ = 2 (v₁ / v₂)

we see the value of the moment depends on the speed of the particles

Answer:

10J

Explanation:

In this question we have the following information

The charge of the particle is q = 5 C

The equipotenetial level is V1 = 5.5 v

and also the

equipotenetial level is V2 = 3.5 v

So we calculate the

work done W=q x (v1-v2)

workdone = 5 x (5.5-3.5)

= 5x2

=10 J

Workdone = 10 J

So we conclude that the workdone on a particle with these information is 10j

Answer:

2.9 cm

Explanation:

Assuming that the rear wheel has a radius of 0.330 m

Given that

r(a) = 12 cm -> 0.12 m

w(a) = 0.6 rev/s -> 3.77 rad/s

v = 5 m/s

r(w) = 0.330 m

The speed on any point on the rim at the sprocket in the front is

v(a) = w(a).r(a) = 3.77 * 0.12 = 0.4524 m/s

Also,

v(a) = speed at any point on the chain

v(b) = speed at any point on the rim of the rear sprocket

v(a) = v(b)

where v(b) = w(b).r(b)

Recall that the speed at any point on the rear wheel is v, where

v = w(b).r(w)

5 = w(b) * 0.330

w(b) = 5/0.330

w(b) = 15.15 rad/s

On substituting this in the equation, we have

v(b) = w(b).r(b).

Remember also, that v(a) = v(b), so

0.4524 = 15.15 * r(b)

r(b) = 0.4524 / 15.15

r(b) = 0.029 m -> 2.9 cm

Therefore, the radius of the rear sprocket needed is 2.9 cm

Answer:1 because

Explanation: it’s pointing to the earth and gravity

Pulls things down to earth

Answer:Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as  

K =-V dP/dV

Where  K, the bulk modulus of water = 2.15 x 10^9N/m^2

2.15 x 10^9N/m^2= - 4,300 x  4 × 106N/m2 / dV

dV = - 4,300 x  4 × 10^6N/m^2/ 2.15 x 10^9N/m^2

dV (change in volume)= -8.000cm^3

Final volume after pressure is applied,

V= V+ dV

V= 4300cm3 + (-8.000cm3)

=4300cm3 - 8.000cm3

Final Volume, V =4,292cm3

Answer:

θ = 122°

Explanation:

Components of a Vector

A vector in the plane can be defined by its rectangular components:

[tex]\vec A =<x,y>[/tex]

Or also can be given by its polar components:

[tex]\vec A =<r,\theta>[/tex]

Where r is the magnitude of the vector and θ is the angle it forms with the positive direction of x.

The relation between them is:

[tex]r=\sqrt{x^2+y^2}[/tex]

[tex]\displaystyle \theta=\arctan\frac{y}{x}[/tex]

It's given the x-component of vector A is x=-25 m and the y-component is y=40 m

(a)

The magnitude of the vector is:

[tex]r=\sqrt{(-25)^2+40^2}[/tex]

[tex]r=\sqrt{625+1600}[/tex]

[tex]r=\sqrt{2225}[/tex]

[tex]r\approx 47.2\ m[/tex]

(b)

[tex]\displaystyle \theta=\arctan\frac{40}{-25}[/tex]

[tex]\displaystyle \theta=\arctan (-1.6)[/tex]

The calculator gives us the value

θ = -58°

But the real angle lies on the second quadrant since x is negative and y is positive, thus:

θ = -58° + 180° = 122°

θ = 122°

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.