PLEASE HELP !
What is the acceleration of a ball that starts at rest and increases in speed
to 20 m/s in 45 seconds? *

Answer:

The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s

Explanation:

Free Fall Motion

A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is [tex]g = 9.8 m/s^2.[/tex]

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The distance traveled by a dropped object is:

[tex]\displaystyle y=\frac{gt^2}{2}[/tex]

Given a grapefruit free falls from a tree and hits the ground t=0.72 s later, we can calculate the height it fell from:

[tex]\displaystyle y=\frac{9.8\cdot 0.72^2}{2}[/tex]

y = 2.54 m

The final speed is computed below:

[tex]vf=9.8\cdot 0.72[/tex]

vf = 7.06 m/s

The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s

Answer:

D

Explanation:

their octet is complete they don't react with any thing

Argon and krypton have similar properties, and they have a complete octet, which means that they do not react with anyone. Hence, option D is correct.

The phrase "inert gas" is a bit misleading because, in some circumstances, these gases can really be reactive. As a result, these gases are typically referred to as noble gases in the context of chemistry and materials science.

The term "noble" has historically been used in chemistry (and earlier in alchemy) to characterize the resistance of some metals to chemical reaction, and the term "noble gas" denotes the same resistance.

These are the inert gases which are mentioned in the periodic table.

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Answer:

c

Explanation:

impulse Is the product of force and distance so it's generally formula

The impulse should be equivalent to the option c.  a force applied to an object for a period of time.

Impulse of force represent the product of the resultant force ΣF and also the duration of the force Δt when the force should be constant. Also it should be lead to the changed to motion and changes to momentum

Therefore, The impulse should be equivalent to the option c.  a force applied to an object for a period of time.

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Answer:

[tex]Force = 13.8N[/tex]

Explanation:

Given

[tex]Mass = 0.6kg[/tex]

[tex]Acceleration = 23m/s^2[/tex]

Required

Determine the applied force

From the question, we understand that no other force acts on the ball.

i.e. the only applied force on the ball is the force applied by the striker.

So, we apply Newton's second law to solve this question.

And this implies that:

[tex]Force = Mass * Acceleration[/tex]

[tex]Force = 0.6kg * 23m/s^2[/tex]

[tex]Force = 13.8N[/tex]

Hence, the applied force by the striker on the ball is 13.8N

The question is incomplete, the complete question is;

A student is measuring the volumes of nectar produced by a flowering plant for an experiment. He measures nectar from 50 flowers using a graduated cylinder that measures to the nearest millilitre (mL). Which statement describes a change that can help improve the results of his experiment?

A.) His measurements will be more precise if he takes measurements from an additional 100 flowers. B.) His measurements will be more accurate if he uses a graduated cylinder that measures to the nearest tenth of a mL. C.) His measurements will be more precise if he uses a graduated cylinder that measures to the nearest tenth of a mL. D.) His measurements will be more accurate if he takes measurements from an additional 100 flowers.

Answer:

His measurements will be more accurate if he uses a graduated cylinder that measures to the nearest tenth of a mL.

Explanation:

In the measurements of volume using most graduated cylinders, the cylinders are calibrated to the nearest tenth owing to the uncertainty in the measurement of volume.

Hence if a cylinder has measures to the nearest milliliter(mL), then he can improve his experiment by using a graduated cylinder that measures to the nearest tenth of a mL

Answer:

The range of the projectile is 60 meters

Explanation:

To determine the range/distance of the projectile, the formula for velocity is used;

velocity = distance/time

where velocity is 15 m/s

time is 4 seconds

distance is unknown

From the formula above, distance is made the subject and thus

distance = velocity × time

distance = 15 × 4

distance = 60 m

The range of the projectile is 60 meters

Answer:

a. 23.9 m/s b. 69.58°

Explanation:

a. Since the space vehicle is moving at a constant velocity of 22.4 m/s in the y direction relative to the space station, its initial vertical velocity v = 22.4 m/s.

Also, since the space vehicle moves in the y direction, its initial horizontal velocity u = 0 m/s.

Since its acceleration a = 0.206 m/s² for time, t = 40.5 s, we find the final horizontal velocity u' from u' = u + at.

Substituting the values of the variables into the equation, we have

u' = u + at

u' = 0 m/s  + 0.206 m/s² × 40.5 s

= 8.343 m/s

≅ 8.34 m/s

The resultant velocity relative to the space station V = √(v² + u'²)

= √[(22.4m/s)² + (8.34 m/s)²]

= √[501.76 m²/s² + (69.56 m²/s²]

= √[571.32 m²/s²]

= 23.9 m/s

b. The direction of the vehicle's velocity relative to the space station is thus θ = tan⁻¹(v/u')

= tan⁻¹(22.4 m/s/8.34 m/s)

= tan⁻¹(2.6959)

= 69.58°

Answer:

In a chemical reaction, only the atoms present in the reactants can end up in the products. No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange amd form new bonds to make the products.

Answer:

99.48

Explanation:

99.48

Answer:

The answer is C

Explanation:

Answer:

The resistance of the lamp is 4Ω.

Explanation:

You have to apply voltage formula :

V = I × R

R = V ÷ I

R = 240 ÷ 60

R = 4 Ω

Answer:100 g of ca

Explanation:

Answer: a. 126.21secs

b. 12.621 meters.

Explanation:

Given data:

Speed of tortoise = 0.1m/s.

Speed of hare = 2m/s.

Solution:

a. Distance traveled = Speed* Time

Speed of tortoise = 0.1 m/s

Speed of hare = 20*0.1 m/s = 2 m/s

2 minutes = 2* 60 s = 120 s

Let the time taken for the race be t seconds.

• Distance moved by tortoise

= (0.1 /s)* (t s)

= 0.1*t meter

•Hare has run for a time of (t - 120)s.

distance moved by hare

= Speed * Time

= (2 m/s)*(t- 120)s

= (2t - 240) meter.

Since hare is 20 cm (0.2 m) behind the tortoise, therefore

(0.1*t - 0.2) meter

= (2t - 240) meter

0.1*t - 0.2 = 2t - 240

Collect like terms

239.8 = 1.9t

Divide both sides by 1.9

t = 126.21secs

The race lasted for 126.21secs

b. Length of race

= Distance moved by tortoise

= 0.1*126.21 meter

= 12.621 meter

The length of the race is 12.621 meters.

Answer:

Explanation:

hope it helps:)

Answer:

Vf = 210 [m/s]

Av = 105 [m/s]

y = 2205 [m]

Explanation:

To solve this problem we must use the following formula of kinematics.

[tex]v_{f} =v_{o} +g*t[/tex]

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (released from the rest)

g = gravity acceleration = 10 [m/s²]

t = time = 21 [s]

Vf = 0 + (10*21)

Vf = 210 [m/s]

Note: The positive sign for the gravity acceleration means that the object is falling in the same direction of the gravity acceleration (downwards)

The average speed is defined as the sum of the final speed plus the initial speed divided by two. (the initial velocity is zero)

Av = (210 + 0)/2

Av = 105 [m/s]

To calculate the distance we must use the following equation of kinematics

[tex]v_{f} ^{2} =v_{o} ^{2} +2*g*y\\\\(210)^{2} = 0 + (2*10*y)[/tex]

44100 = 20*y

y = 2205 [m]

Answer:

Ro = 133 [kg/m³]

Explanation:

In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

[tex]Ro = m/V[/tex]

where:

m = mass [kg]

V = volume [m³]

We will convert the units of length to meters and the mass to kilograms.

L = 15 [cm] = 0.15 [m]

t = 2 [mm] = 0.002 [m]

w = 10 [cm] = 0.1 [m]

Now we can find the volume.

[tex]V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ][/tex]

And the mass m = 4 [gramm] = 0.004 [kg]

[tex]Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}][/tex]

Answer:

7.07 hours

Explanation:

divide the distance by the speed

so in this case, divide 672 by 95

Explanation:

Please mark me as the brainliest answer

Answer:

The electrostatic force is larger by a factor of 8.988E9 / 6.674E-11 = 1.35E20

Explanation:

William gave you all the ingredients, but not the answer.  Being both 1/r^2 forces (resulting from massless mediators in QFT), and with unit charges in the problem (1 coulomb, 1 Kg), separated by unit distance (1m), the only nontrivial numerical values in the problem are the constants of proportionality: Coulomb's constant (k) and Newton's gravitational constant (G).  So to "compare and constrast": the ratio of the forces is simply the ratio of these constants.  The electromagnatic force is 1.35 X 10^20 times stronger than the gravitational force.  Assuming positive charge on both objects (the problem is ambiguous on this), they are repelled, whereas the much weaker gravitational force is attractive.  (Gravity only has one kind of "charge" - it's unsigned - and the force is always attractive).  

The problem doesn't say the objects are pointlike: if they have some extent and are either conductive or made of some dielectric, then things get messy because the charge distribution on the object won't be uniform then, but save that for grad school. :-)

Answer:

Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.

Explanation:

Hope this helps! :]

Answer:

C. Surface 1 is blacktop, Surface 2 is gravel, and

Surface 3 is ice.

Explanation:

Answer:

...........................

Answer:

b

Explanation:

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 200 × 9.8

We have the final answer as

Hope this helps you

Answer:

Explanation:

Step one:

given data

initial speed u= 5m/s

time t= 3seconds

acceleration due to gravity g= 9.8m/s^2

Required:

the final velocity v

Step two:

applying the first equation of motion

v=u+gt-----------we used + because the ball is falling with gravity

v=5+9.8*3

v=5+29.4

v=34.4

The final velocity is 34.4m/s

Answer:

-34.4

Explanation:

the person above had the right approach but it would be negative as it is in a downwards motion

Answer:

In addition to depleting resources for forest-dependent communities, deforestation is contributing to an increase in small-scale natural disasters.“That means when there's heavy rainfall, forest soil can absorb water underground and disburse it to streams.

Answer:   Deforestation invites natural disasters by the following reasons:-

i) Soil erosion

ii)Landslides due to soil erosion

iii) Forest fire

iv) Flooding due to soil erosion

Hope it helped u,

pls mark this as brainliest

and put thanks

^_^

Answer:

Displacement is how far you are between your initial and finishing position.

Distance is how far you went.

Explanation:

For example, using a baseball field... if the sides are each 100 m..

A batter runs to third base... what is his displacement and distance?

Since the batter ran to 3rd base, it means he ran 300 m to get to 3rd base but his displacement is 100 m becuase that's how far they are from home plate. (which was their starting position)

So the answer to my example above would be:

Distance: 300 m

Displacement: 100 m

I hope this helped you.

The correct answer is A physical change

Explanation:

Jorge's experiment shows water at different temperatures; in this experiment, it is expected at low temperatures such as -20°C water is in solid-state (ice), at medium temperatures such as 40°C water is in a liquid state (liquid water), and at high temperatures such as 120°C water is in gaseous state (water vapor). This implies during this experiment the changing factor is the physical state (solid, gas, or liquid), and this is a physical change because only the physical properties of water change but not its composition or identity. According to this, the correct answer is physical change.

Explanation:

s = ut + 1/2 a t^2

200 = 0 * 6 + 1/2 * a * (6)^2

200 = 1/2 * a * 36

200 = 18 a

a = 200/18

a= 11.1m/sec^2

v = u + at

v = 0 + 11.1 * 6

v = 66.6m/s

hope it helps you

Answer:

rs 900

Explanation:

the process is given

Answer:

2917.4 m/s

Explanation:

From the question given above, the following data were:

Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth

Radius (r) of the Moon = 1737 Km

Escape velocity (v) =?

Next, we shall determine the gravitational acceleration of the Moon. This can be obtained as follow:

Gravitational acceleration of the earth = 9.8 m/s²

Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth

= 0.25 × 9.8 = 2.45 m/s²

Next, we shall convert 1737 Km to metres (m). This can be obtained as follow:

1 Km = 1000 m

Therefore,

1737 Km = 1737 Km × 1000 m / 1 Km

1737 Km = 1737000 m

Thus, 1737 Km is equivalent to 1737000 m

Finally, we shall determine the escape velocity of the rocket as shown below:

Gravitational acceleration of the Moon (g) = 2.45 m/s²

Radius (r) of the moon = 1737000 m

Escape velocity (v) =?

v = √2gr

v = √(2 × 2.45 × 1737000)

v = √8511300

v = 2917.4 m/s

Thus, the escape velocity is 2917.4 m/s

Answer:

a

Yes it clears

b

 [tex]b= 0.19 \ m[/tex]

c

 No it does not clear

d

[tex]z= 0.86 \ m[/tex]

Explanation:

From the question we are told that

  The speed at which the player serves the ball is [tex]v = 23.6 \ m/s[/tex]

  The height of the ball above the ground is  [tex]h = 2.3 7 \ m[/tex]

  The distance of the net is  [tex]d = 12 \ m[/tex]

   The height of the net is [tex]H = 0.9 \ m[/tex]

Generally the time taken for the ball to reach the net is mathematically represented as

     [tex]t = \frac{d}{v}[/tex]

=>  [tex]t = \frac{12}{23.6}[/tex]

=>  [tex]t = 0.508 \ s[/tex]

Generally the change in height of the ball after t is mathematically represented as

     [tex]\Delta h = ut + \frac{1}{2} gt^2[/tex]

Here u is the initial velocity which is zero given that the ball was at rest initially

So

     [tex]\Delta h = 0* t + \frac{1}{2} * 9.8 * 0.50 8^2[/tex]

=>  [tex]\Delta h =1.28 \ m[/tex]

Generally the new height of the ball is mathematically evaluated as

      [tex]s= h-\Delta h[/tex]

=>   [tex]s = 2.37 - 1.28[/tex]

=>   [tex]s = 1.09 \ m[/tex]

From the value obtained we see that [tex]s > H[/tex] hence the ball clears the net

Generally the distance between the center of the ball and the top of the net is mathematically represented as

      [tex]b = s - H[/tex]

=>   [tex]b = 1.09 - 0.90[/tex]

=>   [tex]b= 0.19 \ m[/tex]

Given that the ball makes an angle of [tex]5^o[/tex] with the horizontal  , the velocity along the x-axis is  

      [tex]v_x = v cos(5)[/tex]

=>   [tex]v_x = 23.6 cos(5)[/tex]

=>   [tex]v_x = 23.5 \ m/s[/tex]

The velocity along the y-axis is  

      [tex]v_y = v sin(5)[/tex]

=>   [tex]v_y = 23.6 sin(5)[/tex]

=>   [tex]v_y = 2.06 \ m/s[/tex]      

Generally the time taken for the ball to reach the net is

      [tex]t = \frac{d}{v_x}[/tex]

=>   [tex]t = \frac{12}{23.5}[/tex]

=>   [tex]t =0.508 \ s[/tex]

Generally the change in height of the ball after t seconds is  

     [tex]c = v_yt + \frac{1}{2}gt^2[/tex]

=>  [tex]c = 2.06 * 0.508 + \frac{1}{2}* 9.8 * 0.508 ^2[/tex]

=>  [tex]c = 2.33[/tex]

Generally the new height of the ball after time t seconds is  

     [tex]e = h - c[/tex]

=>   [tex]e = 2.37 - 2.33[/tex]

=>   [tex]e = 0.04 \ m[/tex]

From the value obtained we see that [tex]e < H[/tex] hence the ball does not clear the net

Generally the distance between the center of the ball and the top of the net is mathematically represented as

      [tex]z = H-e[/tex]

=>   [tex]z = 0.90 - 0.04[/tex]

=>   [tex]z= 0.86 \ m[/tex]

(a) Yes, the ball clears the net.

(b) The distance between the center of the ball and the top of the net is 0.203 m.

(c) No, the ball does not clear the net.

(d) Now, the distance between the center of the ball and the top of the net is -0.85 m.

When any object or body is launched with some initial velocity and making some angle with the horizontal, the body travels in a parabolic path. It is known as the projectile motion.

Given,

The horizontal distance traveled by the ball is 12 m.

The height of the top of the net is 0.90 m.

The height of the horizontal launch of the ball is 2.37 m.

The time for the horizontal motion of the projectile that is the ball is,

[tex]\begin{aligned} {{v}_{0x}}&={{S}_{x}}t \\ t&=\dfrac{{{S}_{x}}}{{{v}_{0x}}} \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}\cos 0{}^\circ } \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}} \end{aligned}[/tex]

The equation for the vertical motion of the projectile can be solved by substituting the above result.

[tex]\begin{aligned} y&={{y}_{0}}+{{v}_{0y}}t-\frac{1}{2}g{{t}^{2}} \\ y&={{y}_{0}}+\left( {{v}_{0}}\sin 0{}^\circ \right)\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)-\frac{1}{2}g{{\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)}^{2}} \\ \left( 0.90\text{ m}+h \right)&=\left( 2.37\text{ m} \right)+0-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( \frac{12\text{ m}}{23.6\text{ m/s}} \right)}^{2}} \\ h&=2.37\text{ m}-\text{1}\text{.267 m}-0.90\text{ m} \\ &=0.203\text{ m} \end{aligned}[/tex]

Now, consider the case when the ball is thrown with an angle [tex]\bold{5^{\circ}}[/tex] with the horizontal.

The horizontal component of the initial velocity is [tex]\bold{v_0 \cos 5{}^\circ.}[/tex]

The vertical component of the initial velocity is [tex]\bold{v_0 \sin 5{}^\circ.}[/tex]

For the horizontal distance traveled by the ball in this case, the time taken can be calculated as below,

[tex]\begin{aligned} {t}'&=\frac{{{{{S}'}}_{x}}}{{{{{v}'}}_{0x}}} \\ &=\frac{12\text{ m}}{\left( 23.6\text{ m/s} \right)\cos 5{}^\circ } \\ &=0.51\text{ s} \end{aligned}[/tex]

Now, the vertical distance above the ground, y’, traveled by the projectile till reaching the net can be determined as,

[tex]\begin{aligned} {y}'&={{{{y}'}}_{0}}+{{{{v}'}}_{0y}}{t}'-\frac{1}{2}g{{{{t}'}}^{2}} \\ &=0\text{ m}-\left( \left( 23.6\text{ m/s} \right)\sin 5{}^\circ \right)\left( 0.51\text{ s} \right)-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( 0.51\text{ s} \right)}^{2}} \\ &=-1.05\text{ m}-1.28\text{ m} \\ &=-2.32\text{ m} \end{aligned}[/tex]

The height above the top of the net can be determined by adding the above result with (2.37 m - 0.90 m) which is the height of the net’s top relative to the launch position.

[tex]\begin{aligned} {h}'&=\left( 2.37\text{ m}-0.90\text{ m} \right)+{y}' \\ &=\left( 2.37\text{ m}-0.90\text{ m} \right)-2.32\text{ m} \\ &=-0.85\text{ m} \end{aligned}[/tex]

Thus, the distance between the center of the ball and the top of the net is -2,32 m.

When the ball leaves the racquet at 5.00° below the horizontal, the distance between the center of the ball and the top of the net is -0.85 m.

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