The property of an object at rest to remain at rest is known as??

That's if the answer should be in m/s2.

Answer:

b

Explanation:

Answer:

168

Explanation:

56 x 3

have a nice day

Answer:

the answers, the correct one is C,   v₀ₓ = vₓ = 3.91 m / s

Explanation:

This is a projectile launching exercise, in this case they indicate that when leaving the cliff it goes horizontally, therefore the initial vertical speed is zero, let's find the time to reach the base

        y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

at the base the height is zero (y = 0 m)

        0 = y₀ + 0 - ½ g t²

         t = √ (2y₀ / g)

we calculate

        t = √ (2 200 / 9.8)

        t = 6.389 s

with this time we calculate the horizontal speed

        v₀ₓ = x / t

        v₀ₓx = 25 / 6,389

        v₀ₓ = vₓ = 3.91 m / s

When checking the answers, the correct one is C

Answer:

im sorry but what is the question please tell me in the comments s i can help you

Explanation:

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.

Learn more about projectile motion here:

https://brainly.com/question/11049671

Answer:

The ball has no momentum

Explanation:

The given parameters are;

The mass of the ball = 5 kg

The velocity of the ball = 0 (The ball is sitting on the floor without moving)

The momentum of the ball = The mass of the ball × Velocity of the ball

Therefore, the momentum of the ball = 5 kg × 0 m/s = 0

The momentum of the ball is zero, the ball has no momentum.

Answer:

285

Explanation:

read and alot jbgdxtffhgdae

Answer:

i think the answer is #3. :)

Answer:

Explanation:

electrical field in vertical direction = V / .02  , .02 is distance between plate and V is potential difference .

force on electrons = V x e / .02

acceleration of falling electrons

= V x e /( .02  x m )

= 50 x 1.8 x 10¹¹ V

a = 9 x 10¹² V .

Time to cover length of plate

t = .05 / 10⁷ = 5  x 10⁻⁹ s ,      velocity of electron = 10⁷ m /s .

vertical distance covered by electron s = .01 m with acceleration a in time t

s = 1/2 a t²

.01 = .5 x 9  V  x 10¹²  x ( 5 x 10⁻⁹ )²

10⁻² = 4.5 x 25  x 10⁻¹⁸ x 10¹² V

V = 10⁴ /  (4.5 x 25)

= 88.89 Volt .

Explanation:

Given that,

Christina drives her moped 7 kilometers North. She stops for lunch and then drives 5 kilometers East. At first she traveled North and then East. They are perpendicular to each other.

Total path traveled = distance covered by an object

D = 7 km + 5 km = 12 km

Shortest path covered = Displacement

[tex]d=\sqrt{7^2+5^2} \\\\d=8.6\ km[/tex]

For direction :

[tex]\tan\theta=\dfrac{d_y}{d_x}\\\\\tan\theta=\dfrac{7}{5}\\\\\theta=\tan^{-1}(\dfrac{7}{5})\\\\\theta=54.46^{\circ}[/tex]

She is moving in the direction of 54.46 degrees.

Explanation:

because 9 is more than 1.5 and the more water, the more heat

Answer:

i think its 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 = 5 × 2.5

We have the final answer as

Hope this helps you

Answer:

i beleive i cannot help you in physics

Explanation:

Answer:

Explanation:

According to equation of motion;

\sum F = ma

m is the mass of the body

a is the acceleration

\sum F = Fm - Ff (along the horizontal)

Fm is the moving force

Ff is the frictional force

Given;

Fm = 12N

Ff = nR = nmg

Ff = n(4)(9.8)

Ff = 39.2n

Substitute the given value into the expression above;

Fm - Ff = ma

12 - 39.2n = 4(2)

-39.2n = 8-12

-39.2n = -4

n = 4/39.2

n = 0.1

Hence the coefficient of kinetic friction between the table and jug is 0.1

Answer: sonar and radar

Explanation:

Picture

Answer:

The primary colors of light are red, green, and blue. If you subtract these from white you get cyan, magenta, and yellow. ... Mixing these three primary colors generates black. As you mix colors, they tend to get darker, ending up as black.

"The connective tissue that's found between bones."

An example would be between the vertabrae in the spine.

Like the other (inappropriate) answer said, a ligament is what attches muscles and bones.

Also, the tissue inside of bones is bone marrow.

If you know that, you can take out those two options (Option 1 and 3)

Answer:

firm, whitish, flexible connective tissue found in various forms in the larynx and respiratory tract, in structures such as the external ear, and in the articulating surfaces of joints. It is more widespread in the infant skeleton, being replaced by bone during growth

Answer: c living in a camber in an under water habitat

Explanation:

Answer:

Interplanting is the practice of planting a fast-growing crop between a slower-growing one to make the most of your garden space. ... Intercropping enables you to boost the health of all plants because it can enhance soil fertility and cooperation among different plants.

Intercropping refers to large, often commercial, farming operations and is the procedure of planting alternating crop rows. Growing two or more crops in close proximity is known as intercropping.

Starting to grow two or more crops in close vicinity is known as intercropping.

The most common goal of intercropping is to increase yield on a given plot of land by utilizing resources that would otherwise go unused by a single crop.

Intercropping is the cultivation of more than one crop in the same space at the same time. Intercropping is the simultaneous planting of two or more species in a mixture, or the interplanting of one species during the growth of another.

The terms are frequently and regionally used interchangeably. Intercropping is the practice of planting alternating crop rows in large, often advertising, farming operations.

Thus, this is the difference between intercropping and interplanting.

For more details regarding interplanting, visit:

https://brainly.com/question/28875609

#SPJ2

Answer:

NO

Explanation:

I=mr^2 this means moment inertia depends upon mass and square of radius or distance.

Answer:

The resultant displacement of the airplane in 4 hours is 212.8 km.

Explanation:

The components of the airplane's velocity and wind's velocity are:

Airplane:

[tex] v_{a_{x}} = v_{a}cos(45) = 70 km/h*cos(45) = 49.50 km/h [/tex]

[tex] v_{a_{y}} = v_{a}sin(45) = 70 km/hsin(45) = 49.50 km/h [/tex]

Wind:

[tex] v_{w_{x}} = 0 [/tex]

[tex] v_{w_{y}} = v_{w} = -30 km/h [/tex]

Now, to know the new velocity of the airplane we to find the result vector:

[tex] v_{x} = v_{a_{x}} + v_{w_{x}} = 49.50 km/h + 0 = 49.50 km/h [/tex]

[tex]v_{y} = v_{a_{y}} + v_{w_{y}} = 49.50 km/h - 30 km/h = 19.50 km/h[/tex]  

Now, the magnitude of the new speed of the airplane is:

[tex] v_{a} = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(49.50 km/h)^{2} + (19.50 km/h)^{2}} = 53.20 km/h [/tex]

Finally, after 4 hours the resultant displacement of the airplane is:

[tex] x = v*t = 53.20 km/h*4 h = 212.8 km [/tex]

Therefore, the resultant displacement of the airplane in 4 hours is 212.8 km.

I hope it helps you!                                        

potassium ions into the cell and

outside

⇒ sodium ions out of the cell.

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 = 7 × 5

We have the final answer as

Hope this helps you