A 4 kg jug of juice is pulled forward across a horizontal breakfast table, by a 12 N forward-directed force. A frictional force also impedes the jug's motion. What is the coefficient of kinetic friction between the table and jug, if the jug accelerates to the right at 2 m/s^2 ?

A 0.08
B 0.1
C 0.2
D 0.3

Answer:

True

Explanation:

Answer:

285

Explanation:

read and alot jbgdxtffhgdae

potassium ions into the cell and

outside

⇒ sodium ions out of the cell.

Answer:

4,900

Explanation:

you find the weight by multiplying the ball in grams to the gravity

The weight or force due to gravity of a 500 g ball is 4.9 N.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Given in the question  a 500 g ball,

Weight = mass.gravity

Weight = .5*9.8

Weight = 4.9 N

The weight or force due to gravity of a 500 g ball is 4.9 N.

To learn more about force refer to the link:

brainly.com/question/13191643

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Answer: C.) Efficiency

Explanation: Edge 2020 good luck ^__^

Answer:

The moment of a given force about a given axis of rotation can be decreased by decreasing the perpendicular distance of force from the axis of rotation.

Answer:

hope it helps...

Explanation:

The moment of a given force about a given axis of rotation can be decreased by decreasing the perpendicular distance of force from the axis of rotation.

Answer:

What field(s) does a moving electron have?

Explanation:

magnetic and gravitational fields

For Edmentem!

Answer: mass = 10 kg, acceleration = 3 m/s2.

Explanation: I hope that helped!!

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 .

Answer:

i beleive i cannot help you in physics

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 = 7 × 5

We have the final answer as

Hope this helps you

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!                                        

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

"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:

i think the answer is #3. :)

Answer:

i think its b

Explanation:

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.

Answer:

The equation Pi = c/2r means that Pi (r) = c/2, so r = c / (2 Pi). The radius of a circle is therefore equal to c / (2 Pi), where c is the circumference of the circle.

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:

E, F, A, C

Explanation:

They cannot change the way a trait is observed

Mutations could change protein shape(A), could cause genetic disorders and harm you(E), could actually be beneficial and cause adaptations to your environment(F), or they could be neutral in which no changes are observantly notable(C).

Hope this helps

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

Explanation:

If the box weighs 100N on Earth, then its mass is 10.2 kg. A pulling force has nothing to do with it.

Explanation:

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

Answer:

The reason why it is important that all the airplanes have the same design is that the students first test is an experiment to determine the effect of the paper kind on the distance the airplane flies

The first experiment, therefore measures the effect of the independent variable, which is the kind of paper on the dependent variable, which is how far the paper flies

The constant or the factor held constant in the experiment is the design of the airplane, so as to ensure that changes in the distance flown by the airplane are due to the changes in the kind of paper alone

Explanation:

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

Answer:

b

Explanation: