The mass of a dropped object impacts its final velocity but not its acceleration.

Answer:With gravity, two things with mass will want to move toward each other. However, we humans don't feel our gravity pulling on another person because it's not very big, but we do all feel the pull of Earth's gravity all the time - we're not all floating in the air, because that would be happening without Earth's gravity!

Explanation:

Answer:

x=4953.6m

Explanation:

used formula x=xo+vot+1/2at^2

Answer:

The value is    [tex]t = 56.68 \  s  [/tex]

Explanation:

From the question we are told that

   The rating of the hoist motor is  [tex]k  =  159hp = 159 *746 =118614 \ W[/tex]

    The  percentage of it power used is  [tex]z = 0.72 * 118614=85402.08 \ W[/tex]

      The  mass of the load is m  = 5550 kg

      The distance is  h = 89.0 m

The potential  energy required to lift the load through that distance is

     [tex]E =  m *  g * h[/tex]

=>    [tex]E =  5550 *  9.8 *  89.0[/tex]

=>   [tex]E =  4840710 \ J[/tex]

Generally the time taken is mathematically represented as

       [tex]t = \frac{E}{ z}[/tex]

=>    [tex]t = \frac{4840710}{ 85402.08}[/tex]

=>    [tex]t = 56.68 \  s  [/tex]

Answer:

1 Newton

Explanation:

Mass of bowl (Mb) = 50 g = 0.05 kg

Initial Velocity (Vo) = 10 mL/s = 10g/s

time (t) = 5.0s

Mass of water (Mw) = 10 g/s * 5s = 50 g = 0.05 kg

*Vob = initial velocity of bowl

*Vow = initial velocity of water

*Vf = final velocity

*g = gravity = 9.8

Conservation of momentum [Inelastic Collision]:

Impulse-Momentum Theorem:

Scale Reading:

Answer:

apex consumers

Explanation:

they are top

Answer:

im sure its A

Explanation:

Answer:

superset

Explanation:

Complete Question

The acceleration of the spacecraft in which the Apollo astronauts took off from the moon was 3.4 m/s2. On the moon, g = 1.6 m/s2.  What was the apparent weight of a 75 kg astronaut during takeoff?

Answer:

The value is  [tex]N = 375 \ N[/tex]

Explanation:

From the question we are told that

  The acceleration is  [tex]a = 3.4 \ m/s^2[/tex]

   The acceleration due to gravity in the moon is [tex]g = 1.6 m/s^2[/tex]

   The mass of the astronaut is  [tex]m = 75 \ kg[/tex]

Generally the apparent weight is mathematically represented as

      [tex]W = ma + mg[/tex]

=>    [tex]W = 3.4 * 75 + 1.6 * 75[/tex]

=>    [tex]W = 375 \ N[/tex]

Answer:speeding up constantly

Explanation:

The graph between the time and the speed of the car shows that the speed is increasing constantly, so, option C is correct.

A moving object's rate of change in distance traveled is measured as speed. Speed is a scalar, which implies it is a measurement with a magnitude but no direction.

A thing that moves quickly and with high speed, covering a lot of ground in a short time. On the other hand, a slow-moving object traveling at a low speed covers a comparatively small distance in the same amount of time. An object with zero speed does not move at all.

Given:

The graph shows the speed of a car traveling east over a 12-second period,

As you can see from the graph, at time t = 0 sec the speed is 10 m/s,

At t = 3 sec, the speed = 15.3 m/s

At t = 6 sec, the speed = 20.3 m/s

Thus, speed is increasing constantly.

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Answer:

A window

Explanation:

Answer:

a window.. duh

Explanation:

Answer:

options C is correct

Explanation:

asking questions is super in this education life

Answer:

option c should be the answer

Answer:

A Cell

Explanation:

Answer:

Explanation:

wavelength λ = 580 x 10⁻⁹ m

slit width d = 3.7 x 10⁻⁶ m

distance of screen D = 10 m

distance of first bright fringe =  1.5 x λ D / d

= 1.5 x 580 x 10⁻⁹  x 10 / 3.7 x 10⁻⁶

= 2351.34  x 10⁻³ m

= 2351.34 mm .

Answer:

The Speed of the vehicles is 9.34m/s

Explanation:

For an elastic collision the two bodies move with similar velocities after collision

Given

M1=1330kg

V1=15m/s

M2=805kg

V2=0(the car is parked on neutral)

The formula is

M1V1+M2V2=(M1+M2)V

1330*15+805*0=(1330+805)V

19950+0=2135V

2135V=19950

divide both sides by 2135

V=19950/2135

V=9.34m/s

Answer:

it takes more effort to lift

Explanation:

because your moving the object weights up. While pushing, you only have to over ride the friction of the object against the ground. Because the coefficient of friction between a box and the ground is usually less than 1.

Answer:

Divide by 1000

Explanation:

From the calculations, the resistance of the material is 1.1 ohm.

Given that;

R α l/A

R = ρl/A

R = resistance

l = length

A = Area

ρ = resistivity

Now;

A = πr^2

A = 3.142 * (1.5 * 10^-3/2)^2

A = 1.77 * 10^-6 m^2

ρ = RA/l

ρ =  2.5 *  1.77 * 10^-6 /20

ρ = 2.2 * 10^-7 ohm/m

Now;

R =ρl/A

A =  3.142 * (3  * 10^-3/2)^2

A = 7.1 * 10^-6 m^2

Thus

R = 2.2 * 10^-7 * 35/ 7.1 * 10^-6

R = 1.1 ohm

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Answer:

The average force exerted on the car during this time is 5,943 N

Explanation:

Given;

mass of the car, m = 849 kg

initial velocity of the car, u = 0

time of motion of the car, t = 5.00 s

final velocity of the car, v = 35 m/s

The average force exerted on the car during this time is given by;

[tex]F = ma \\\\F = \frac{m(v-u)}{t}\\\\F = \frac{849(35-0)}{5}\\\\F = \frac{849(35)}{5}\\\\ F = 849*7\\\\F = 5,943 \ N[/tex]

Therefore, the average force exerted on the car during this time is 5,943 N

Answer:

5943N

Let's say (+x) = eastward

Average horizontal acceleration

ax = vx -v0x/5.00s

= 35.0m/s-0/5.00s

= +7.09m/s

From here we apply the second law of newton

During this period average horizontal force acting on car

Summation x = max = (849kg)(+7.09m/s²)

= 5943N

+5.943x10³N

= 5.94kN east ward.

Answer:

300

Explanation:

450Newton × 2Meter ÷ 3sec

Answer:

I think it's = 12 seconds

Explanation:

the formula for speed is:

speed=[tex]\frac{distance}{time}[/tex] SO, time is equal to:

time=[tex]\frac{distance}{speed}[/tex]

(sub the numbers in the formula)

distance=1200m, speed=100m/s

time=[tex]\frac{1200}{100}[/tex]

=12 seconds

Answer:

m = 50 [kg]

Explanation:

In order to solve this problem we must be clear about the difference between weight and mass. Weight is the product of mass by the acceleration of the planet or the star. While the mass is always preserved it never changes regardless of where it is located.

So for the earth we have:

g = gravity acceleration = 9.8 [m/s^2]

m = mass [kg]

W = weigth = 490 [N]

therefore the mass will be:

m = W/g

m = 490/9.8

m = 50 [kg]

Now it is important to remember that the mass will be the same on the moon or on the earth, but the weight will be different, because the gravity acceleration of the moon is different from the gravity acceleration on earth

So the gravity on the moon is equal to:

81.7 = 50 * gm

gm = 1.634 [m/s^2]

Answer:

R u from k12?? i am XD add me on discord my username is Hot Boy#1650

Explanation:

We are given:

Mass of object (m) = 12 kg

acceleration (a) = 6 m/s²

Solving for the Force:

From newton's second law of motion:

F = ma

replacing the variables

F = 12*6

F = 72N

Answer:

It will rotate counter-clockwise.

Explanation:

The reason is that there is more Nm on the left side which will lift the lever towards the left side.

The lever would rotate counterclockwise as the torque on the left is higher, so, option B is correct.

The force which causes the object to rotate about any axis is called perpendicular distance.

Torque is a twisting or turning force that frequently results in rotation around an axis, which may be a fixed point or the center of mass. The ability of something rotating, such as a gear or a shaft, to overcome turning resistance is another way to think of torque.

Given:

The force on the left side, f = 40 N,

The force on the right side, F = 80 N,

The distance of the pivot from the left side, d = 9 m,

The  distance of the pivot from the right side, d = 3 m,

Calculate the torque on both side as shown below,

Torque on the left side = 40 × 9

Torque on the left side = 360 Nm

Torque on the right side = 80 × 3

Torque on the right side = 240 Nm,

Here, the torque on the left side is more,

Thus, the lever will rotate counterclockwise.

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Answer:

A)  t = 7.25 sec

B) 45.92 m above the cliff and 85.92 m counting from ground level

C) about 0.6 m/s

D) The only acceleration is always that of gravity : 9.8 m/s^2

Explanation:

The kinematic equation for the position as a function of time is:

y = 40 + 30 * t - (g/2) t^2

The total time the balloon was in the air is calculated for when the final position is y = 0 (when it touches the ground), that is:

0 = 40 + 30 * t - (g/2) t^2

0 = 40 +30 t - 4.9 t^2

which can be solved for "t" using the quadratic formula, and which renders two different times "t", of which we pick the positive answer: t = 7.25 sec.

For the maximum height, we estimate the time it takes for the balloon to reach the maximum height at which the velocity is zero (changes direction of motion):

vf = vi - g * t

0 = 30 - 9.8 * t

t = 30 / 9.8 = 3.06 seconds

Now we use this value in our position equation and get:

ymax =  40 + 30 (3.06) - 4.9 (3.06)^2

ymax = 85.92 m  (from ground level

Therefore, from the cliff (that is at 40 m height) the height is 85.92 - 40 = 45.92 m.

The velocity of the balloon at t = 3 seconds can be calculated with the velocity expression we have been using:

v(t) = vi - g * t

v(3) = 30 - 9.8 (3) = 0.6  m/s

The only acceleration acting on the balloon is always the acceleration due to gravity (g = 9.8 m/s^2)

Answer:

1. [tex] E = 1.14 \cdot 10^{13} N/C [/tex]

2. [tex]E_{k} = 9.1 \cdot 10^{-17} J[/tex]      

Explanation:

1. The strength of the nucleus' electric field (E):

[tex]E = \frac{kq}{r^{2}}[/tex]

Where:

k: is the Coulomb constant = 9x10⁹ Nm²/C²

q: is the proton charge = 1.6x10⁻¹⁹ C

r: is the radius = 10⁻¹⁰ m

[tex]E = \frac{kq}{r^{2}} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*79*1.6 \cdot 10^{-19} C}{(10^{-10} m)^{2}} = 1.14 \cdot 10^{13} N/C[/tex]

2. The kinetic energy (Ek) of an electron is the following:

[tex] E_{k} = \frac{1}{2}mv^{2} [/tex]    

Where:

m is the electron's mass = 9.1x10⁻³¹ kg

v: is the speed of the electron

We can find the speed of the electron by equaling the centripetal force (Fc) and the electrostatic force (Fe):

[tex] F_{c} = F_{e} [/tex]  

[tex] \frac{mv^{2}}{r} = \frac{kq^{2}}{r^{2}} = qE [/tex]

[tex] v^{2} = \frac{qEr}{m} = \frac{1.6 \cdot 10^{-19} C*1.14 \cdot 10^{13} N/C*10^{-10} m}{9.1 \cdot 10^{-31} kg} = 2.00 \cdot 10^{14} m^{2}/s^{2} [/tex]                  

Now, we can find the kinetic energy:

[tex] E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}9.1 \cdot 10^{-31} kg*2.00 \cdot 10^{14} m^{2}/s^{2} = 9.1 \cdot 10^{-17} J [/tex]    

I hope it helps you!