If you placed a negatively charged object within this electric field, which direction will it move?

Final answer:

Use the kinematic equation for projectile motion, S = ut + 0.5at², to find the time it takes for a stone thrown upward at 352 feet per second to hit the ground considering the acceleration due to gravity as 32.2 feet per second squared. Calculate the time to the peak, and then double it to find the total time.

Explanation:

The student has asked about the time it will take for a stone thrown upward at an initial speed of 352 feet per second to hit the ground. To find the time it takes for the stone to hit the ground, we can use the kinematic equations of motion that describe the motion of projectiles under the influence of gravity. Since the acceleration due to gravity is 32.2 feet per second squared downward, we can use the equation  S = ut + 0.5at², where S is the displacement (which would be zero, as the stone returns to the ground level), u is the initial velocity, a is the acceleration due to gravity, and t is the time.

We rearrange the equation to solve for t. However, we need to consider that the stone will take as much time to reach the highest point as it will take to fall back down from that point. Therefore, we calculate the time to rise to the peak using the equation u = gt, and then double it. Plugging in the values, we get t = u/g, which gives us t = 352/32.2, and doubling this time gives us the total time to hit the ground.

Answer: Correct answer is "The marbles horizontal velocity changes, but its vertical velocity does not."

Explanation: I have had the same question on my High School Physics test, and got the answer correct.

Formed from melting glaciers

Most lakes are formed from melting glaciers.

The box is moving at constant speed, so acceleration=0 and the equation of the forces acting on the box along the ramp becomes:

[tex]F-F_f-W_p = 0[/tex] (1)

where F=200 N is the force that pushes the box, Ff is the frictional force, and Wp is the component of the weight parallel to the ramp.

The frictional force can be written as

[tex]F_f = \mu m g cos \theta[/tex]

where [tex]\mu=0.18[/tex] is the coefficient of friction, m is the mass of the box, [tex]g=9.8 m/s^2[/tex] is the frictional force and [tex]\theta=30^{\circ}[/tex] is the angle of the ramp.

The component of the weight along the ramp is

[tex]W_p = mg sin \theta[/tex]

Substituting this into eq.(1), we have

[tex]F-\mu mg cos \theta- mg sin \theta=0[/tex]

and we can find the mass of the box:

[tex]m=\frac{F}{\mu g cos \theta + g sin \theta}=\frac{200 N}{(0.18)(9.8 m/s^2)(cos 30^{\circ})+(9.8 m/s^2)(sin 30^{\circ})}=31.1 kg[/tex]

Now we can also find the work done on the box. this is given by the gain in potential energy of the box, since there is no change in kinetic energy (the speed of the box is constant). Since the box has risen vertically by 1 m, the gain in potential energy (and the work done) is

[tex]W=\Delta U=mg \Delta h=(31.1 kg)(9.8 m/s^2)(1 m)=304.8 J[/tex]

So, the two answers are

(a) 304.8 J

(b) 31.1 kg

The physics concept of work and forces were used to answer this question. The work done was calculated to be 200 Joules and the mass was determined using the given force, angle, and coefficient of friction.

Let's solve this problem using the concepts of physics. First, we need to find how much work was done. Work done (W) = Force (F) x Distance (d) x cos (theta), where theta is the angle between the force and the direction of displacement. Here, as the force was directed up the slope and the box moved up the slope too, the angle (theta) is 0 degrees. Cos 0 = 1, so the equation reduces to W = F x d.

Given, Force (F) = 200 N and Distance (the vertical height) (d) = 1 m (Remember, the work done is equal to the force multiplied by the distance moved in the direction of the force). So, the work done = 200 N x 1 m = 200 Joules.

Next, we need to determine the mass of the box. The forces acting on the body will be gravity (mg), friction (fr) and the applied force (F). According to the information given, the box moved up the slope at a constant speed. This means the net force acting on the box should be zero. Therefore, F - fr - mg*sin(theta) = 0. Given is the coefficient of friction as 0.18, so fr = mu*N = mu*mg*cos(theta), where N = mg*cos(theta) is the normal force. Substituting fr into the net force equation, we can solve for mass (m) = F/(g*(mu*cos(theta) + sin(theta))). Substituting the known values, we get the mass of the box.

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circle graph

Explanation:

Answer:

Reducing the normal force against the ground will reduce the amount of friction force, you can do this reducing the weight or mass of the object that you want to move.

Smoothing down the surface, for example, on a wet floor, the water fills some of the little imperfections in the ground, so the coefficient of dynamic friction in the ground is reduced.

Answer:

Work done, W = 3200 ergs

Explanation:

It is given that

Force acting on the object, F = 0.01 N

Distance covered by the object, d = 0.032 m

Let W is the work done by the object. The product of force and the distance is called the work done by the object. Mathematically, it is given by :

[tex]W=F\times d[/tex]

[tex]W=0.01\ N\times 0.032\ m[/tex]    

W = 0.00032 Joules

Since, [tex]1\ joule=10^7\ ergs[/tex]

This gives, work done by the object, W = 3200 ergs. So, the correct option is (a).                                                          

Explanation :

It is given that :

Initial velocity, [tex]v_0=50\ m/s[/tex]

Launching angle, [tex]\theta_0=36.9^0[/tex]

The motion followed by this object is called as its projectile motion.

the horizontal component is given by:

[tex]x_t=x_0+v_0cos\theta_0t[/tex].......(1)

Put all values in equation (1)

For t = 0, [tex]x_t=0[/tex]

For t = 1 s, [tex]x_1=0+50\ cos(36.9)\ 1=39.95\ m[/tex]

For t = 2 s, [tex]x_2=0+50\ cos(36.9)\ 2=79.9\ m[/tex]

For t = 3 s, [tex]x_3=0+50\ cos(36.9)\ 3=119.85\ m[/tex]

For t = 4 s, [tex]x_3=0+50\ cos(36.9)\ 4=159.8\ m[/tex]

For t = 5 s, [tex]x_4=0+50\ cos(36.9)\ 5=199.75\ m[/tex]

For t = 6 s, [tex]x_3=0+50\ cos(36.9)\ 6=239.7\ m[/tex]

Hence, this is the required solution.

Answer:d

Explanation:

Final answer:

The atomic nucleus, located at the center of an atom, consists of positively charged protons and neutral neutrons. Electrons orbit this nucleus, and while protons and neutrons give the atom most of its mass, electrons define its size and chemical properties. Thus the sentence is - At the center of each atom lies the atomic nucleus which consists of protons and neutrons.

Explanation:

At the center of each atom lies the atomic nucleus which consists of protons and neutrons.

The nucleus is a positively charged region at the center of the atom. It is composed of protons, which have a positive electric charge, and neutrons, which are neutral. This nucleus is very dense and contains the vast majority (99.97%) of an atom's mass. Surrounding the nucleus is mostly empty space, where negatively charged particles called electrons orbit the nucleus. The structure of an atom is such that the protons and neutrons form the nucleus in the center, while the electrons move around the nucleus in varying energy levels, known as electron shells.

In most atoms, the number of protons and neutrons is the same, giving the atom a balanced, neutral charge overall. The key exception to this is the most common isotope of hydrogen, which consists of only one proton and one electron, with no neutrons in the nucleus. Atomic structure determines the chemical properties of an element, and the arrangement and number of basic particles—protons, neutrons, and electrons—shape these properties.

So, what's the answer ?

Answer: The terminal velocity of the skydiver is 281.30 m/s.

Explanation:

Mass of the sky diver = 75 kg

Density of the air at height of 39,000 m =4.3% of the surface density of the air

= [tex]1.2 kg/m^3\times \frac{4.3}{100}=0.0516 kg/m^3[/tex]

Cross sectional area of the sky diver , A = [tex]0.72 m^2[/tex]

Coefficient of drag ,C = 0.5

The terminal speed is given by the formula:

[tex]v_T=\sqrt{\frac{2mg}{c\rho A}}[/tex]

[tex]v_T=\sqrt{\frac{2\times 75 kg\times 9.8 m/s^2}{0.5\times 0.0516 km/m^3\times 0.72 m^2}}=281.30 m/s[/tex]

The terminal velocity of the skydiver is 281.30 m/s.

The terminal speed of the skydiver is [tex]\boxed{281.452{\text{ m/s}}}[/tex].

Further Explanation:

When a body falls from the sky it will accelerate continuously in the downward direction by the force of gravity results in increment in the acceleration. Another force is acting on the body is friction force due to air or the particles present in the atmosphere in the in the upward direction or opposite to the direction of motion also called drag force. The drag force is directly proportional to the square of the velocity of falling body. Therefore, with increase in the velocity the drag force also increasing which is proportional to equal to the square of the velocity of falling body. And time will come where the drag force is equal to the weight or the force of gravity results in zero resultant force means zero acceleration. At this point the speed become constant and reached its maximum velocity. This speed is called the terminal velocity.

Given:

The height of the jump is [tex]39000\text{ m}[/tex].

The density air is [tex]4.3\%[/tex] of the surface density.

The density of air at sea level is [tex]1.2{\text{ kg/}}{{\text{m}}^3}[/tex].

The mass of the skydiver is [tex]75\text{ kg}[/tex].

The cross sectional area of the skydiver is [tex]0.72{\text{ }}{{\text{m}}^2}[/tex].

Concept:

The expression for the terminal speed is:

[tex]\fbox{\begin\\{v_t} =\sqrt {\dfrac{{2mg}}{{\rho A{C_d}}}}\end{minispace}}[/tex]                                                                                                                                                           …… (1)

Here, [tex]{v_t}[/tex] is the terminal speed, [tex]m[/tex] is the mass of the skydiver, [tex]g[/tex] is the acceleration due to gravity, [tex]\rho[/tex]  is the density, [tex]A[/tex] is the cross sectional area and [tex]{C_d}[/tex] is the drag coefficient.

The term [tex]g[/tex] is called acceleration due to gravity and it has a constant value of [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex] and the drag coefficient of a human in free fall is [tex]0.5[/tex].

The density of the air at the altitude of [tex]39000\text{ m}[/tex] is:

[tex]\rho=4.3\%\times{\rho _{air}}[/tex]

Here,[tex]\rho[/tex]  is the density of the air at the altitude of [tex]39000\text{ m}[/tex] and [tex]{\rho _{air}}[/tex] is the surface density of the air.

Substitute [tex]1.2{\text{ kg/}}{{\text{m}}^3}[/tex] for [tex]{\rho _{air}}[/tex] in the above equation.

[tex]\begin{aligned}\rho&=4.3\%\times1.2{\text{ kg/}}{{\text{m}}^3}\\&=\frac{{4.3}}{{100}}\times1.2{\text{ kg/}}{{\text{m}}^3}\\&=0.0516{\text{ kg/}}{{\text{m}}^3}\\\end{aligned}[/tex]

Substitute [tex]0.0516{\text{ kg/}}{{\text{m}}^3}[/tex] for [tex]\rho[/tex], [tex]75\text{ kg}[/tex] for [tex]m[/tex], [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex] for [tex]g[/tex], [tex]0.72{\text{ }}{{\text{m}}^2}[/tex] for [tex]A[/tex] and [tex]0.5[/tex] for [tex]{C_d}[/tex] in equation (1).

[tex]\begin{aligned}{v_t}&=\sqrt{\frac{{2\left({75{\text{ kg}}}\right)\left( {9.81{\text{ m/}}{{\text{s}}^2}} \right)}}{{\left({0.0516{\text{ kg/}}{{\text{m}}^3}}\right)\left({0.72{\text{}}{{\text{m}}^2}} \right)\left( {0.5} \right)}}}\\ &=\sqrt{\frac{{1471.5}}{{0.018576}}}{\text{ m/s}}\\&=\sqrt{79215.116}{\text{ m/s}}\\&=281.452{\text{ m/s}}\\ \end{aligned}[/tex]

Therefore, the terminal speed of the skydiver is [tex]\boxed{281.452{\text{ m/s}}}[/tex].

Learn more:

1. Terminal velocity https://brainly.in/question/6499953

2. Terminal velocity https://brainly.com/question/8789089

3. Net force https://brainly.com/question/11799308

Answer Details:

Grade: High school

Subject: Physics

Chapter: Kinematics

Keywords:

Air, less, dense, higher, elevation, skydivers, reach high, terminal speed, recorded, achieved, jump, height, 39,000 m, density, air, 4.3%, surface density, estimate, sea, level,  , mass, 75 kg, cross section, area,  , 281.452 m/s.

Answer:

The correct answer is B) Fewer peaches will be sold.

Explanation:

Hope this helps! :)

The false statement about the tundra is D. The tundra biome does indeed support animal life, including polar bears, caribou, and mountain goats, among others.

The correct answer to which of the following statements about the tundra is not true is D. The tundra biome does not support animal life is the false statement. Although the tundra environment is extreme, the biodiversity, while limited, does include a variety of animal life that has adapted to the harsh conditions.

Arctic tundra and alpine tundra are the two main categories of the tundra biome. The arctic tundra can go below -20° F and has limited vegetation, primarily mosses, lichens, short grasses, and some flowering plants. The alpine tundra is found at high altitudes with a longer summer growing season compared to the arctic tundra.

Animals such as polar bears, caribou, and mountain goats, as well as birds that migrate to the arctic tundra in the summer, are adapted to survive in the tundra biome. These animals have special adaptations to handle the cold temperatures and scarce resources found in the tundra.

Answer:

To answer this question, we need to recur to the law of inertia.

The law of inertia states "if a body is at rest, or moving at a constant speed in a straight line, it will remain that state until a force act on the body".

So, in this case, the jet is traveling at a speed of 600 km/h, which means that all bodies inside the jet are also traveling at this speed, because they are in it.

Also, if we apply the law of inertia here, imagine that somebody falls from the jet, that person will travel at a speed near to 600km/h, if we discard the wind, following a constant trajectory.

The same phenomenon happens with the pillow, when it falls from an overhead rack, if the passanger is passing under the trajectory of the pillow, it will slam her, because due to intertia, the pillow will continue its movement until something stop it, which can be the person's body.