Which answer is it
0.32 m/s
3.2 m/s
9.8 m/s
40 m/s

Answer:

Approximately [tex]-15.37m/s^2[/tex]

Explanation:

The acceleration can be found using the formula:

[tex]x_{f} = x_{i} + v_{i} (t)+\frac{1}{2} (a)(t^{2} )[/tex]

The work is as shown:

[tex]45m = 0 + 39 m/s (3.3s)+\frac{1}{2} a(3.3s)^2[/tex]

[tex]45m=128.7m + 5.445s^2 a[/tex]

[tex]-83.7m=5.445s^2a[/tex]

[tex]a = -15.37190083m/s^2[/tex]

a is about -15.37[tex]m/s^2[/tex]

Answer:

The final speed of the electron is 6.626 x 10⁷ m/s

Explanation:

force applied to the electron, f = 4 x 10⁻¹⁵ N

distance traveled by the electron, d = 50 cm = 0.5 m

The work done on the electron;

W = F x d

W = (4 x 10⁻¹⁵ N) x (0.5 m)

W = 2 x 10⁻¹⁵ J

The kinetic energy of the electron is given by;

[tex]K.E = \frac{1}{2}m_ev^2[/tex]

Apply work - energy theorem;

K.E = W

[tex]\frac{1}{2}m_ev^2 = 2 *10^{-15}\\\\ v^2 = \frac{2( 2 *10^{-15})}{m_e}\\\\v= \sqrt{\frac{2( 2 *10^{-15})}{m_e}}\\\\ v = \sqrt{\frac{2( 2 *10^{-15})}{9.11*10^{-31}}}\\\\v = 6.626*10^7 \ m/s[/tex]

Therefore, the final speed of the electron is 6.626 x 10⁷ m/s

Answer:

The object on the moon has greater mass

Explanation:

Given;

weight of the object on Earth, W₁ = 1 N

wight of the object on Moon, W₂ = 1 N

gravity on Earth, g₁ = 9.8 m/s²

gravity on the Moon, g₂ = g₁/6 = 1.633 m/s²

Apply Newton's second law of motion;

Weight of the object on Earth is given by;

W₁ = m₁g₁

m₁ = W₁/g₁

m₁ = 1 / 9.8

m₁ = 0.102 kg

Weight of the object on Moon is given by;

W₂ = m₂g₂

m₂ = W₂ / g₂

m₂ = 1 / 1.633

m₂ = 0.612 kg

Therefore, the object on the moon has greater mass.

Answer:

Explanation:

Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.

Answer:

Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.

Explanation:

Answer:

0.2533333334

Explanation:

It is simple, just do the math

Answer:

Wnet = 0

Explanation:

        [tex]W_{net} = \frac{1}{2} k* \Delta x^{2} - m*g*h_{max} (1)[/tex]

        [tex]\frac{1}{2} k* \Delta x^{2} = m*g*h_{max} (2)[/tex]

Angular velocity of the coil :

[tex]\omega=\dfrac{2\pi n}{60}[/tex]

[tex]\omega=\dfrac{2\times \pi\times 3500}{60}\\\\\omega =366.52\ rad/s[/tex]

Now,

Peak voltage is given by :

[tex]V=NAB\omega\\\\V= 200\times \dfrac{\pi (0.1)^2}{4}\times 0.7\times 366.52\\\\V=403.01\ V[/tex]

Hence, this is the required solution.

Answer:

It is increasing .60 m/s

Explanation:

Answer:

The question is incomplete. However the complete question is :

A Rail Gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basic mechanism of acceleration is relatively simple and can be illustrated in the following example. A metal rod of mass m and electrical resistance R rests on parallel horizontal rails (that have negligible electric resistance), which are a distance L apart. The rails are also connected to a voltage source V, so a current loop is formed.

The vertical magnetic field, initially zero, is slowly increased. When the field strength reaches the value [tex]$B_0$[/tex], the rod, which was initially at rest, begins to move. Assume that the rod has a slightly flattened bottom so that it slides instead of rolling. Use g for the magnitude of the acceleration due to gravity.

Find μs, the coefficient of static friction between the rod and the rails.

Express the coefficient of static friction in terms of variables given in the introduction.

So the answer is : [tex]$\mu_s=\frac{\Delta V LB_0}{Rmg}$[/tex]

Explanation:

It is given : B = [tex]$B_0$[/tex]

Therefore, using force on the current carrying wire is:

[tex]$\mu_s mg = B_0IL$[/tex]

[tex]$\mu_s= \frac{ILB_0}{mg}$[/tex]

Therefore,

[tex]$\mu_s=\frac{\Delta V LB_0}{Rmg}$[/tex]

The coefficient of static friction in terms of variables will be [tex]\rm \mu_s =\frac{ \triangle VLB_0}{Rmg} \\\\[/tex].

The ratio of the greatest static friction force (F) between the surfaces in contact before movement begins to the normal (N) force is the coefficient of static friction.

The given data in the problem is;

[tex]\rm B=B_0[/tex]

The force on a current-carrying wire is balanced by the friction force.

The force on a current-carrying wire is given by;

[tex]\rm \mu_s mg= B_0IL \\\\ \rm \mu_s=\frac{ILB_0}{mg}\\\\ \rm \mu_s= \frac{\triangle VLB_0}{Rmg}[/tex]

Hence the coefficient of static friction in terms of variables will be [tex]\rm \mu_s =\frac{ \triangle VLB_0}{Rmg} \\\\[/tex].

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Answer:

0.04558

Explanation:

Luminosity is defined as the total amount of energy a star can produce in a second. Luminosity is given by the formula

σ * A * T⁴

where;

σ = Stefan-Boltzmann constant

= 5.670367 * 10⁻⁸

A= Surface Area

=4πr²

T= Temperature in Kelvin

In the question,

Temperature= 4000K

Radius= 2.0 A.U

Substituting the variables into the equation,

0.000000056704 * 4(3.14)(2.0²) * 4000⁴

= 0.000000056704 * 50.24 * 16000

= 0.04558

Answer:

answer is C

Explanation:

Answer: Monetary policy used during times of recession

Explanation:

Answer:

Ax = -3       By= 7         tan phi =  -7 / 3    phi = -66.8

Since theta is in the second quadrant   theta = 180 - 66.8 = 113.2 deg

Bx = 5        By = 2        tan theta = .4        theta = 21.8 deg

Cx = Ax + Bx = 2       Cy = Ay + By = 9

tan theta = 9 / 2 = 4.5      theta = 77.5 deg

1) The solution is [tex]\theta_{1} \approx 113.141^{\circ}[/tex].

2) The solution is [tex]\theta_{1} \approx 21.874^{\circ}[/tex].

3) The solution is [tex]\theta_{1} \approx 77.467^{\circ}[/tex].

1) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:

[tex]\cos \theta = \frac{\vec A\,\bullet \, \vec u}{\|\vec A\|\cdot \|\vec u\|}[/tex] (1)

Where [tex]\|\vec A\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec A[/tex] and [tex]\vec u[/tex].

If we know that [tex]\vec A = (-3, 7)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:

[tex]\|\vec A\| = \sqrt{(-3)^{2}+7^{2}}[/tex]

[tex]\|\vec A\| = \sqrt{58}[/tex]

[tex]\|\vec u\| = 1[/tex]

[tex]\cos \theta = \frac{(-3)\cdot (1)}{\sqrt{58} \cdot 1}[/tex]

[tex]\cos \theta \approx -0.393[/tex]

The solution is [tex]\theta_{1} \approx 113.141^{\circ}[/tex].

2) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:

[tex]\cos \theta = \frac{\vec B\,\bullet \, \vec u}{\|\vec B\|\cdot \|\vec u\|}[/tex] (1)

Where [tex]\|\vec B\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec B[/tex] and [tex]\vec u[/tex].

If we know that [tex]\vec B = (5, 2)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:

[tex]\|\vec B\| = \sqrt{5^{2}+2^{2}}[/tex]

[tex]\|\vec B\| = \sqrt{29}[/tex]

[tex]\|\vec u\| = 1[/tex]

[tex]\cos \theta = \frac{(5)\cdot (1)}{\sqrt{29} \cdot 1}[/tex]

[tex]\cos \theta \approx 0.928[/tex]

The solution is [tex]\theta_{1} \approx 21.874^{\circ}[/tex].

3) In this case, we need to find the value of [tex]\vec C[/tex] by vectorial sum before calculating the angles with respect to x-axis:

[tex]\vec C = \vec A + \vec B = (-3, 7) + (5, 2)[/tex]

[tex]\vec C = (2, 9)[/tex]

In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:

[tex]\cos \theta = \frac{\vec C\,\bullet \, \vec u}{\|\vec C\|\cdot \|\vec u\|}[/tex] (1)

Where [tex]\|\vec C\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec C[/tex] and [tex]\vec u[/tex].

If we know that [tex]\vec B = (2, 9)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:

[tex]\|\vec C\| = \sqrt{2^{2}+9^{2}}[/tex]

[tex]\|\vec C\| = \sqrt{85}[/tex]

[tex]\|\vec u\| = 1[/tex]

[tex]\cos \theta = \frac{(2)\cdot (1)}{\sqrt{85} \cdot 1}[/tex]

[tex]\cos \theta \approx 0.217[/tex]

The solution is [tex]\theta_{1} \approx 77.467^{\circ}[/tex].

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Answer:

B.11.3 kg

Explanation :

F=ma here force is given as 17N and acceleration as 1.5 m/sec/sec using the formula mass can be calculated

∴ F=ma

17N =m × 1.5 m[tex]s^{-2}[/tex]

[tex]m=\frac{F}{a}[/tex]

[tex]m= \frac{17 kg m/sec/sec}{1.5 m/sec/sec}[/tex]

∴ m=11.3 kg

Answer:

mk7000000.0000000000

Answer:

2.54*10^8 m/s

Explanation:

Given that

Speed of light in space, c = 299792458 m/s

Wavelength of the beam of light, λ = 621 nm or 621*10^-9 m

Index of refraction of the liquid, n = 1.18

Speed of light in the liquid, v = ?

The index of refraction of an optical material, n is given by the following formula.

n = c/v, and if we substitute directly, we have

1.18 = 299792458 / v, if we make v subject of formula, we have

v = 299792458 / 1.18

v = 2.54*10^8 m/s

Answer:

ans: a biological function

Answer:

B

Explanation:

EDGE2020

Answer:

[tex]B=3.29\ \mu T[/tex]

Explanation:

Given that,

Length of a wire is 3 m

Current in the wore, I = 10 A

The wire formed a semicircle.

We need to find the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.

The magnetic field at the center of semicircle is given by :

[tex]B=\dfrac{\mu_o I}{4R}[/tex]

R is radius of semicircle,

[tex]R=\dfrac{3}{\pi}=0.954\ m[/tex]

So,

[tex]B=\dfrac{4\pi \times 10^{-7}\times 10I}{4(0.954)}\\\\B=3.29\times 10^{-6}\ T\\\\B=3.29\ \mu T[/tex]

So, the magnitude of the magnetic field is [tex]3.29\ \mu T[/tex].

Answer:

sciencephysicsphysics questions and answersA 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. ...

Question: A 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. The Block Is Released From Rest And Slides Across A Track As Shown In (Figure 1). While Most Of The Track Is Frictionless, There Is A 55-cm Section Of Track That Has A Coefficient Of Friction With The Block Of 0.4. A Bit Further On, The Track Ascends ...

A 12-kg block is pressed against a spring (spring constant 870 N/m ), compressing it some distance. The block is released from rest and slides across a track as shown in (Figure 1). While most of the track is frictionless, there is a 55-cm section of track that has a coefficient of friction with the block of 0.4. A bit further on, the track ascends into a hill that is 40-cm tall.

What is the minimum compression of the spring necessary for the block to slide past the section with friction?

What is the minimum compression of the spring necessary for the block to make it to the top of the hill on the other side? 40cm 12kg Ww.w.wwwww 55cm

Explanation:

Answer:

Explanation:

Let force P = 193 pounds

second force = Q

Angle that resultant makes with force P is θ

Tanθ = Q sin86.27 / (P + Q cos86.27 )

Tan 31.4 = Q sin86.27 / (193 + Q cos86.27 )

.61 = .99 Q / (193 + .065Q )

117.73 + .04 Q = .99Q

117.73 = .95 Q

Q = 123.93  pounds .

Answer:

canoe = 3 h 40 min.  motorboat = 1 h 40 min

Explanation:

        [tex]v_{b} = \frac{\Delta x_{b} }{\Delta t_{b} } (1)[/tex]

       [tex]v_{mb} = \frac{\Delta x_{mb} }{\Delta t_{mb} } (2)[/tex]

        [tex]v_{b} *\Delta t_{b} = v_{mb} *\Delta t_{mb} (3)[/tex]

        [tex]v_{b} *t_{b} = v_{mb} *(t_{b} - 2 hs.) = 10 km/h* t_{b} = 22 km/h* (t_{b} - 2 hs.)[/tex]

        [tex]t_{b} = \frac{-44 }{-12} hs = 3h 40 min.[/tex]

        tmb = tb - 2 hs. = 1h 40 min.

Answer:

D. V/3.

Explanation:

V = rw

At point a

Linear velocity

= Va = rw

At point b

We have,

V = 3rw

So linear velocity of a

= V/Va = 3rw/re

I cross multiples and after which I got Va = v/3

Please view attachment

Answer:

Explanation:

The relation between linear and angular velocity,

[tex]V = rw[/tex]

Now, for point A the linear velocity is

[tex]V_A = rw[/tex]

And for point B

[tex]V = 3rw[/tex]

Therefore, the linear velocity of the point A

[tex]\frac{V}{V_A} = \frac{3rw}{rw}\\\\V_A = \frac{V}{3}[/tex]

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Answer:

E = 1.76 J

Explanation:

Given that,

Mass of an object, m = 0.4 kg

It moves by a vertical distance of 0.45 m in the Earth's gravitational field.

We need to find the change in its gravitational potential energy. It can be given by the formula as follow :

[tex]E=mgh\\\\E=0.4\times 9.8\times 0.45\\\\E=1.76\ J[/tex]

So, the change in its gravitational potential energy is 1.76 J.

Answer:

μ = 0.41

Explanation:

[tex]\frac{1}{2} kx^2 = \mu mgd[/tex]

[tex]\mu = (\frac{1}{2} kx^2)/mgd[/tex]

[tex]\mu = (\frac{1}{2} (100)(0.2)^2)/(0.5)(9.81)(1)[/tex]

[tex]\mu = 0.41[/tex]

The coefficient of kinetic friction between the block and the tabletop is 0.41.

The given parameters;

The coefficient of kinetic friction between the block and the tabletop by applying the principle of conservation of energy as shown below.

Fd = Uₓ

μmgd = ¹/₂kx²

where;

The coefficient of kinetic friction between the block and the tabletop;

[tex]\mu_k = \frac{kx^2}{mgd} \\\\\mu_k = \frac{100\times 0.2^2}{2\times 0.5\times9.8 \times 1 } \\\\\mu_k = 0.41[/tex]

Thus, the coefficient of kinetic friction between the block and the tabletop is 0.41.

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Answer:

Explanation:

A 52.0 kg skateboarder wants to just make it to the upper edge of a "quarter pipe," a track that is one-quarter of a circle with a radius of 2.60 m .

What speed does he need at the bottom

Let the lowest point of the circle = (0,0)

Center of circle = (0, 2.6)

The height of the circle = 2 * radius = 2 * 2.60 = 5.20 m

Highest point = (0, 5.2)

As the skateboarder moves around ¼ of a vertical circle, the skateboarder moves from the lowest position to a position that is ½ the way up to the highest position. This is the point that is 2.6 meters directly to the right of left of the center = (2.6, 2.6)

As the skateboarder has moved 2.6 m upward, the potential energy increase = m * g * ∆h = 52.0 * 9.8 * 2.6

During this same time, the kinetic energy decreased from the maximum to 0. The decrease of KE = the increase of PE

½ * m * ∆v^2 = m * g * ∆h

½ * 52 * ∆v^2 = 52.0 * 9.8 * 2.6

½ * ∆v^2 = 9.8 * 2.6

∆v = (2 * 9.8 * 2.6)^0.5 = 7.14 m/s

The velocity at highest point, (2.6,2.6) is 0 m/s

So the velocity at the lowest point must be 7.14 m/s

Let’s see if the skateboarder has enough KE to move upward 2.6 m.

KE at bottom = ½ * 52 * 7.14^2 = 1325.5

PE = 52 * 9.8 * h

52 * 9.8 * h = 1325.5

h = 2.6 m

The answer is OK

The speed needed by the skateboarder at the bottom is 7.8 m/s.

The given parameters;

Apply the principle of conservation of mechanical energy to determine the sped of the skateboarder at the bottom;

[tex]P.E_{top} = K.E_{bottom}\\\\mgh = \frac{1}{2}mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 3.1} \\\\v = 7.8 \ m/s[/tex]

Thus, the speed needed by the skateboarder at the bottom is 7.8 m/s.

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Answer:

Locating misplaced electrical equipment.

Explanation:

Ultrasound is primarily used by doctors to locate fractured bones or damaged organs, so I assume it can be used to locate other things too.

Answer:

These two objects hit the ground at the same time. But their impact velocity will be different

because they are dropped from different heights.

The one that is dropped from the building will have more impact when it hits the ground because gravity causes this onject to fall toward the ground at a faster and faster velocity the longer the object falls. In fact, its velocity increases by 9.8 m/s2.

Explanation:

Initial velocity of the object u=0

Final velocity of the object when it reaches the ground v=A m/s

Using v^2−u^2=2gS where g=10m/s^2

A^2 −0^2=2(10)H

A^2=20*H H- heigh from the table

Object falling from the building will have the same initial velocity but different heigh.

T>H

so

B^2=20*T

20*T>20*H

so B^2>A^2

it equal to B>A so object falling from the building will have more impact than falling from the table.

Answer:

b the apple changed color

Answer:

0.176

Explanation:

(5.27)/(3.05*9.8)

It is indeed 0.176 for the answer.