Based on the information in the table, what
is the acceleration of this object?

t(s) v(m/s)
0.0
9.0
1.0
4.0
2.0
-1.0
3.0
-6.0
A. -5.0 m/s2
B. -2.0 m/s2
C. 4.0 m/s2
D. 0.0 m/s2

Answer:

A(predicts an organisms ability to adapt to its enviroment, it is not a fact that each organization can adapt)

Explanation:

[tex]\implies {\blue {\boxed {\boxed {\purple {\sf { \frac{1}{I} = \frac{R}{E} + \frac{r}{E} }}}}}}[/tex]

[tex]\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]

[tex]I = \frac{ E}{ R + r} \\[/tex]

[tex] ➺\:\frac{I}{1} = \frac{E}{R + r} \\[/tex]

Since [tex]\frac{a}{b} = \frac{c}{d} [/tex] can be written as [tex]ad = bc[/tex], we have

[tex]➺ \: I \: (R + r) = E \times 1[/tex]

[tex]➺ \: \frac{1}{I} = \frac{R + r}{E} \\ [/tex]

[tex]➺ \: \frac{1}{I} = \frac{R}{E} + \frac{r}{E} \\ [/tex]

[tex]\boxed{ Hence\:proved. }[/tex]

[tex]\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}[/tex]

Explanation:

static equilibrium means its on the floor or something

so slightly greater than 98 newtons in the upward direction

Answer:

v = 87.46 m/s

Explanation:

The radial acceleration is the centripetal acceleration, whose formula is given as:

[tex]a_c = \frac{v^2}{r}[/tex]

where,

[tex]a_c[/tex] = centripetal acceleration = 17 m/s²

v = planes's speed = ?

r = radius of path = 450 m

Therefore,

[tex]17\ m/s^2 = \frac{v^2}{450\ m}\\\\v^2 = (17\ m/s^2)(450\ m)\\\\v = \sqrt{7650\ m^2/s^2}[/tex]

v = 87.46 m/s

Answer:

The simplified expression is  3.833 x 10⁷ g

Explanation:

Given expression;

3.88 x 10⁷ g  -  4.701 x 10⁵ g

The expression above is simplified as follows;

= (3.88 x 100 x 10⁵ )g  -  ( 4.701 x 10⁵) g

= (388 x 10⁵ )g -  ( 4.701 x 10⁵) g

= (388  - 4.701 ) x (10⁵ )g

= 383.299 x 10⁵ g

In standard form, the simplified expression can be expressed as;

= (3.83299 x 100 x 10⁵) g

= 3.83299 x 10⁷ g

= 3.833 x 10⁷ g

Therefore, the simplified expression is  3.833 x 10⁷ g

Answer:

b)

Explanation:

GCMs (general circulation models) are useful instruments for gaining a quantitative knowledge of climate processes. Physical processes in the atmosphere, cryosphere, and land surface are represented by them. They are used for modeling the global climate system's reaction to rising greenhouse gas concentrations available at the moment by utilizing spectral models based on the energy emitted by the biosphere and clouds.

Answer:

Riley's reasoning is correct

Explanation:

Her reasoning is correct because as the ball moves upwards, the acceleration due to gravity would be vertical and in downward position. Therefore at all points as the ball moves, the velocity of the ball is going to change in magnitude as well as in direction. given that the direction keeps changing at certain points, the angle made by the initial velocity just as the ball left the surface would also have to continuously change.

If Riley has to wait for this ball to move some inches before she uses the protractor to measure the angle, the angle of travel would have to change.

Therefore there is going to be discrepancies between the measured angle and the predicted angle. The predicted is the angle of velocity with the horizontal just as this ball moves from the surface.

Answer:

Resistance is inversly proportional to the current.

V=I.R.

P=V.I

Answer:

Option C. 0.73 g/cm³

Explanation:

From the question given above, the following data were obtained:

Mass = 80 g

Area (A) = 10000 cm²

Thickness = 0.11 mm

Density =?

Next, we shall convert 0.11 mm to cm. This can be obtained as follow:

10 mm = 1 cm

Therefore,

0.11 mm = 0.11 mm × 1 cm / 10 mm

0.11 mm = 0.011 cm

Thus, 0.11 mm is equivalent to 0.011 cm.

Next, we shall determine the volume of the paper. This can be obtained as follow:

Area (A) = 10000 cm²

Thickness = 0.011 cm

Volume =?

Volume = Area × Thickness

Volume = 10000 × 0.011

Volume = 110 cm³

Finally, we shall determine the density of the paper. This can be obtained as follow:

Mass = 80 g

Volume = 110 cm³

Density =?

Density = mass / volume

Density = 80 / 110

Density = 0.73 g/cm³

Therefore the density of the paper is 0.73 g/cm³

Let A be the position of the block at the top of the first incline; B its position at the bottom of the first incline; C its position at the bottom of the second incline; and D its position at the top of the second incline. I'll denote the energy of the block at a given point by E (point).

At point A, the block has total energy

E (A) = (10.0 kg) (9.80 m/s²) (7.00 m) + 1/2 (10.0 kg) v₀²

E (A) = 686 J + 1/2 (10.0 kg) v₀²

At point B, the block's potential energy is converted into kinetic energy, so that its total energy is

E (B) = 1/2 (10.0 kg) v₁²

The block then slides over the horizontal surface with constant speed v₁ until it reaches point C and slides up a maximum height of 26.0 m to point D. Its total energy at D is purely potential energy,

E (D) = (10.0 kg) (9.80 m/s²) (26.0 m) = 2548 J

Throughout this whole process, energy is conserved, so

E (A) = E (B) = E (C) = E (D)

(a) Solve for v₀ :

686 J + 1/2 (10.0 kg) v₀² = 2548 J

==>   v₀ ≈ 19.3 m/s

(b) Solve for v₁ :

1/2 (10.0 kg) v₁² = 2548 J

==>   v₁ ≈ 22.6 m/s

Now if the horizontal surface is not frictionless, kinetic friction will contribute some negative work to slow down the block between points C and D. Check the net forces acting on the block over this region:

• net horizontal force:

∑ F = -f = ma

• net vertical force:

∑ F = n - mg = 0

where f is the magnitude of kinetic friction, a is the block's acceleration, n is the mag. of the normal force, and mg is the block's weight. Solve for a :

n = mg = (10.0 kg) (9.80 m/s²) = 98.0 N

f = µn = 0.500 (98.0 N) = 49.0 N

==>   - (49.0 N) = (10.0 kg) a

==>   a = - 4.90 m/s²

The block decelerates uniformly over a distance 2.00 m and slows down to a speed v₂ such that

v₂² - v₁² = 2 (-4.90 m/s²) (2.00 m)

==>   v₂² = 490 m²/s²

and thus the block has total/kinetic energy

E (C) = 1/2 (10.0 kg) v₂² = 2450 J

(c) The block then slides a height h up the frictionless incline to D, where its kinetic energy is again converted to potential energy. With no friction, E (C) = E (D), so

2450 J = (10.0 kg) (9.80 m/s²) h

==>   h = 25.0 m

(d) At half the maximum height, the block has speed v₃ such that

2450 J = (10.0 kg) (9.80 m/s²) (h/2) + 1/2 (10.0 kg) v₃²

==>   v₃ ≈ 15.7 m/s

The block loses speed and thus energy as it moves between B and C, but its energy is conserved elsewhere. If we ignore the inclines and pretend that the block is sliding over a long horizontal surface, then its velocity v at time t is given by

v = v₁ + at = 22.6 m/s - (4.90 m/s²) t

The block comes to a rest when v = 0 :

0 = 22.6 m/s - (4.90 m/s²) t

==>   t ≈ 4.61 s

It covers a distance x after time t of

x = v₁t + 1/2 at ²

so when it comes to a complete stop, it will have moved a distance of

x = (22.6 m/s) (4.61 s) + 1/2 (-4.90 m/s²) (4.61 s)² = 52.0 m

(e) The block crosses the rough region

(52.0 m) / (2.00 m) = 26 times

Answer:

Explanation:

Moment of inertia of solid sphere = 2/5 m R²

m is mass and R is radius of sphere.

Putting the values

Moment of inertia of solid sphere I₁

Moment of inertia of hollow  sphere I₂

Kinetic energy of solid sphere ( both linear and rotational )

= 1/2 ( m v² + I₁ ω²)                [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/5 m R² ω²)

= 1/2 ( m v² + 2/5 m v²)

=1/2 x 7 / 5 m v²

= 0.7 x 5 x 4² = 56 J .

This will be equal to work to be done to stop it.

Kinetic energy of hollow sphere ( both linear and rotational )

= 1/2 ( m v² + I₂ ω²)  [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/3 m R² ω²)

= 1/2 ( m v² + 2/3 m v²)

=1/2 x 5 / 3 m v²

= 0.833 x 5 x 4² = 66.64 J .

This will be equal to work to be done to stop it.

Answer:

Upthrust on boat becomes lesser than Weight of boat

Explanation:

When there are more people than the capacity, The weight of the boat acting downwards increases. However, the upthrust acting on the submerged part of the boat is constant. Since Weight > Upthrust, there is a net force downwards, leading to sinking.

Answer:

the rock speed is increased

A rock is initially at rest concerning the earth, but the speed of the rock will increase when it hits the ground.

The resistance to something rolling or moving over another solid object is called friction. Even though frictional forces can be helpful, like the traction needed to walk without slipping, they also present a considerable amount of resistance to motion. About 20% of the engine power in a car is used to combat frictional forces in the moving parts.

The primary cause of friction between metals appears to be the forces of attraction, also known as adhesion, between the contact zones of the surfaces, which are always microscopically unequal. Because of friction caused by the imperfections of the tougher surface rubbing up against the softer surface, these "welded" connections are sheared.

To know more about Friction :

https://brainly.com/question/13000653

#SPJ2

Answer:

5.04 m

Explanation:

You are told that the homeowner wants to increase their fences by 34 percent meaning Original+ 34 percent. If the original is 100 percent, then the new fence size will be 134 % of the original. You are given the original which is 3.76 meters, to find new fence size 1.34 * 3.76m to get 5.0384 meters, rounded to 5.04 m.

Answer:

5.0384m

Explanation:

% increase = 100 x (Final - Initial / | initial | )

( |~~| Bars indicate absolute value since you can't have a negative height)

Answer:

3.05×10⁵ Nm²C⁻¹

Explanation:

According to Gauss' law,

∅' = q/e₀............... Equation 1

Where ∅' = net flux through the surface, q = net charge, e₀ = electric permittivity of the space

From the question,

Given: q = 2.7 μC = 2.7×10⁻⁶ C,

Constant: e₀ = 8.85×10⁻¹² C²/N.m²

Substituting these values into equation 1

∅' = (2.7×10⁻⁶)/(8.85×10⁻¹²)

∅' = 3.05×10⁵ Nm²C⁻¹

The question is incomplete. The complete question is :

A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].

Determine the following:

(a) frequency of the motion

(b) period of the motion

(c) amplitude of the motion

(d) first time after t = 0  that the object reaches the position x = 2.6 cm

Solution :

Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].

Comparing it with the general equation of simple harmonic motion,

 x = A sin (ωt + Φ)

  A = 4.7 cm

  ω = 7.9 π

a). Therefore, frequency, [tex]$f=\frac{\omega}{2 \pi}$[/tex]

                                             [tex]$=\frac{7.9 \pi}{2 \pi}$[/tex]

                                             = 3.95 Hz

b). The period, [tex]$T=\frac{1}{f}$[/tex]

                        [tex]$T=\frac{1}{3.95}[/tex]

                            = 0.253 seconds

c). Amplitude is A = 4.7 cm

d). We have,

    x = A sin (ωt + Φ)

    [tex]$x_t=4.7 \sin (7.9 \pi t)$[/tex]

    [tex]$2.6 = 4.7 \sin (7.9 \pi t)$[/tex]

     [tex]$\sin (7.9 \pi t) = \frac{26}{47}$[/tex]

     [tex]$7.9 \pi t = \sin^{-1}\left(\frac{26}{47}\right)$[/tex]

          Hence, t = 0.0236 seconds.

Answer:

The plane will be located directly above the bomb because they both have the same horizontal speed.

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

            x ’= 368.61 m

            y ’= 450 sin 35

            y ’= 258.11 m

Answer:

[tex]I_2=30.9A[/tex]

Explanation:

From the question we are told that:

Wire segment [tex]l_s=2.9m[/tex]

Initial Current [tex]I_1=1400A[/tex]

Force [tex]F=2.00N[/tex]

Distance of Wire [tex]d=4.50cm=>0.0450m[/tex]

Generally the equation for Force is mathematically given by

 [tex]F=\frac{\mu_0 * I_1*I_2*l_s}{2 \pi *r}[/tex]

 [tex]F=\frac{4 \pi*10^{-7} *1400 I*I_2*2.9}{2 \pi *0.0450}[/tex]

 [tex]I_2=\frac{22.5*10^-2}{2*10^{-7}*1400*2.6}[/tex]

 [tex]I_2=30.9A[/tex]

Answer:

Explanation:

Distillation is the separation of multiple LIQUID components based on their different BOILING POINT. As the mixture is heated and the first component SEPARATES, its PURE form travels through the distillation set-up and GOES into a different container

Answer:

True

Explanation:

With the gravitational pull that our planets have, we are able to remain in orbit. This demonstrates how every object in the galaxy is attracted to every other object. Every object in the universe that has mass exerts a gravitational pull on every other mass. We as humans do it too, but since our force isn't strong, we don't have much of an effect. I hope this helped and please don't hesitate to reach out with more questions!

Answer:

elastic energy

Explanation:

When a gymnast falls on a trampoline from a height, after coming in contact with the trampoline, both the gymnast and the trampoline start to move down due to the elastic property of the trampoline.

During this stretching of the trampoline there comes a maximum point up to which the trampoline is stretched. At this point, both the kinetic energy and the gravitational potential energy of the gymnast are zero due to zero speed and zero height, respectively.

The only energy stored in the gymnast's body at this point is the elastic potential energy due to stretching of the trampoline. Hence,the correct option is:

elastic energy

N = 52.0 N

Explanation:

Given: [tex]F_a= 20.0\:\text{N}=\:\text{applied\:force}[/tex]

[tex]m=6.00\:\text{kg}[/tex]

[tex]N = \text{normal force}[/tex]

The net force [tex]F_{net}[/tex] is given by

[tex]F_{net} = N + F_a\sin 20 - mg=0[/tex]

Solving for N, we get

[tex]N = mg - F_a\sin 20[/tex]

[tex]\:\:\:\:\:\:= (6.00\:\text{kg})(9.8\:\text{m/s}^2) - (20.0\:\text{N}\sin 20)[/tex]

[tex]\:\:\:\:\:\:= 52.0\:\text{N}[/tex]

Answer:

(i)8m/s²(ii)40m/s

Explanation:

according to the formula

½at²=s.

then substituting the data

½a•5²=100

a=8m/s²

v=at=8•5=40m/s

Answer:

(I)

[tex]{ \bf{s = ut + \frac{1}{2} a {t}^{2} }} \\ 100 = (0 \times 5) + \frac{1}{2} \times a \times {5}^{2} \\ 200 = 25a \\ { \tt{acceleration = 8 \: m {s}^{ -2} }}[/tex]

(ii)

[tex]{ \bf{v = u + at}} \\ v = 0 + (8 \times 5) \\ { \tt{final \: velocity = 40 \: m {s}^{ - 1} }}[/tex]

Answer:

True. A permanent magnet like the earth produces its own B field due to movement of the iron core. The earths magnetic field is the reason why we have an atmosphere and it also is the only defense against solar flares. A coil of wire or solenoid that has current have so much moving charge that the motion of the electrical charge can create a significant G b-field

Answer:

B) 1500m/s

Explanation:

Ans is 1500m/s

Answer:

Both stones will land at the same time because both stones will fall with the same acceleration through the same height.

Explanation:

We are given that

Mass of stone ,m1=5 Kg

Mass of stone, m2=1 kg

We have to find which stone more faster will land and why.

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

Initial velocity of both stones=0

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

[tex]t^2=\frac{h}{g}[/tex]

[tex]t=\sqrt{\frac{h}{g}}[/tex]

[tex]t_1=t_2=\sqrt{\frac{h}{g}}[/tex]

Because both stones are thrown from the same height.

Both stones will land at the same time because both stones will fall with the same acceleration through the same height  and the acceleration  does not depend of its mass.