A student kicks a soccer ball upward at a 30º angle with an initial speed of 20 m∕s. What expression should the student use to calculate the magnitude of the ball’s initial velocity in the horizontal direction?

Answer:

[tex]72.12\ \text{A/m}^2[/tex]

south

cross sectional area of the beam

Explanation:

v = Velocity of ions = [tex]4.6\times 10^5\ \text{m/s}[/tex]

Number of ions per [tex]\text{cm}^3[/tex] = [tex]4.9\times 10^8[/tex]

Charge density would be the product of number of ions per [tex]cm^3[/tex] and the charge of electrons multiplied by 2 as they are doubly charged.

[tex]\rho_q=4.9\times 10^8\times 10^6\times 2\times 1.6\times 10^{-19}\\\Rightarrow \rho_q=0.0001568\ \text{C/m}^3[/tex]

Current density is given by

[tex]J=\rho_qv\\\Rightarrow J=0.0001568\times 4.6\times 10^5\\\Rightarrow J=72.12\ \text{A/m}^2[/tex]

The current density is [tex]72.12\ \text{A/m}^2[/tex]

The direction of the current density is opposite to the movement of the charged particle. The particles are moving north so the direction of current density will be to the south.

Current is given by

[tex]I=JA[/tex]

where A is the cross sectional area of the beam .

So the cross sectional area of the beam is required to determine the total current in this ion beam.

Answer:

option c is correct

Explanation:

we know that

2as=vf^2-vi^2

vf=24 m/s

vi= 0 m/s

a=g= 9.8 m/s^2

s=vf^2-vi^2/2a

s=(24)²-(0)²/2*9.8

s=576/19.6

s=29.4 m

therefore option c is correct

Answer:

k=320N/m

Explanation:

Step one:

given data

Let the initial/equilibrum position be x

mass m1= 0.2kg

F1= 0.2*10= 2N

elongation e= 9.5cm= 0.095m

mass m2=1kg

F2=1*10= 10N

elongation e= 12cm= 0.12m

Step two:

From Hooke's law, which states that provided the elastic limits of a material is not exceeded the extention e is proportional to applied Force F

F=ke

2=k(0.095-a)

2=0.095k-ka----------1

10=k(0.12-a)

10=0.12k-ka----------2

solving equation 1 and 2 simultaneously

   10=0.12k-ka----------2

-   2=0.095k-ka----------1

   8=0.025k-0

divide both side by 0.025

k=8/0.025

k=320N/m

Answer:

A

Explanation:

Answer:

320 paper clips

Explanation:

mass = volume × density = 20cm³ × 8g/cm³ = 160g

mass of 1 paper clip = 0.50g

mass of x paper clips = 160g

x = 160/0.50 = 320

Answer:

I AM SO SORRY

but here is the answer

Explanation:

20 j

Answer:

Explanation:

The mass of the object can be found by using the formula

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

f is the force

a is the acceleration

From the question we have

[tex]m = \frac{135}{25} = \frac{27}{5} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

16+15+19= ??

Am just messign with u lol

Explanation:

anwser s 19 inches

i

Answer:

hi

Explanation:

how to convert gumbauan baboy to kung burger p

Answer:

ke = 1/2mv^2

    = 1/2 * 1200 * 22^2

    = 600 * 484

    = 290400 J

Hope it helped u,

pls mark this as brainliest

and put thanks

^_^

Answer:

The kinetic energy at this point 290,400

Answer:

a. 18.375m/s²

b. 1.142s

Explanation:

A. Using v² = u² + 2as

Where;

V = final velocity (m/s)

u = initial velocity (m/s)

a = acceleration (m/s²)

s = distance (m)

In this question, v = 21m/s, u= Om/s, s = 12m

21² = 0² + 2 × a × 12

441 = 0 + 24a

441 = 24a

a = 18.375m/s²

B). Using V = u + at

21 = 0 + 18.375t

21 = 18.375t

t = 21/18.375

t = 1.142s

Answer:

Explanation:

F=kx

x=F/k

F=2000 kg

x=100 cm=9*10^-3

effective spring constant=k=F/x

k=2000/9*10^-3=2.2*10^-5

now frequency

f=1/2π√k/m

f=1/2*3.14√2.2*10^-5/310

f=1/6.28√7.097*10^-8

f=1/6.28*2.7*10^-4

f=0.16*2.7*10^-4

f=4.32*10^-5

The effective spring constant of the springs is 33755.55 N/m.

The frequency of the car's vibration is 2.07 Hz.

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

A spring balance can be used to calculate the Force. The Newton is the SI unit of force.

Weight of the four people: F = 310 × 9.80 N = 3038 Newton.

The additional compression of the spring: x = 0.90 cm = 0.90 × 10⁻² m.

Hence, the effective spring constant of the springs: k= force/compression

= 3038 N/0.90 × 10⁻² m

= 33755.55 N/m.

The frequency of the car's vibration is: f = 1/2π√(k/m)

=1/2π√(33755.55/2000)

= 2.07 Hz.

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#SPJ2

Answer:

atoms that have the same # of protons but a DIFFERENT # of NEUTRONS

Explanation:

Answer:

D-They have different number of neutrons

Explanation:

brainliest? Plz

Answer:

The head

Explanation:

Answer:

C. Equals

Explanation:

Law of reflection Equals the angle of incidence

Answer:

D. The balloons will separate further as the repulsion magnitude increases.

Answer:

The time it takes the car to stop is 8.64 s.

Explanation:

Given;

initial velocity of the car, u = 37.3 m/s

final velocity of the car, v = 8.8 m/s

constant acceleration of the car, a = -3.3 m/s²

The time it takes the car to stop is given by;

[tex]t = \frac{v-u}{a}[/tex]

Substitute the givens;

[tex]t = \frac{v-u}{a}\\\\t = \frac{8.8-37.3}{-3.3}\\\\t = 8.64 \ s[/tex]

Therefore, the time it takes the car to stop is 8.64 s.

Answer:

a) 126 kgm/s

b) 0 kgm/s

c) 3.9 m/s

Explanation:

To solve this question, we will use the law of conservation of momentum.

Momentum before collision = momentum after collision

m1v1 + m2v2 = (m1 + m2)v, where

m1 = mass of the shopping cart, 21 kg

m2 = mass of the box, 11 kg

v1 = initial velocity of the shopping cart, 6 m/s

v2 = initial velocity of the box, 0 m/s

v = final velocity of the box+cart

a)

Momentum of the shopping cart before collision = P

P = mv

P = 21 * 6

P = 126 kgm/s = c

b)

Momentum of the box before collision

Like in question a above, the momentum of the box is P

P = mv

P = 11 * 0

P = 0 kgm/s = b

c)

Velocity of the combined shopping cart wreckage after collision is

m1v1 + m2v2 = (m1 + m2)v

(21 * 6) + (11 * 0) = (21 + 11)v

126 + 0 = 32v

32v = 126

v = 126/32

v = 3.9375 m/s, on approximating to 1 decimal place, we have 3.9 m/s and option b as the answer.

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Answer: • They have same change in velocity but different changes in kinetic energy

•They started at the same height.

Explanation:

First and foremost, we need to note that both balls have thesame acceleration due to gravity and due to this, even though they've different masses, they'll fall at same speed.

Also, since kinetic energy that's, the energy relating to motion of a mass, us dependent on mass and speed, their kinetic energy will be different.

Therefore, based in the explanation, the correct options are:

• They have same change in velocity but different changes in kinetic energy

•They started at the same height.

Explanation:

the formula of speed is distance traveled by time it work

31.25 cm/sec

__________________________________________________________

We are given:

Distance travelled = 150 cm

Time taken = 4.8 seconds

Final Speed of the Marble:

Speed of the marble = Distance travelled / Time taken

Speed of the marble = 150 / 4.8

Speed of the marble = 31.25 cm/sec

Answer:

[tex]{:}\longrightarrow[/tex][tex]\sf 1km=1000×100cm [/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf 1km=100000cm [/tex]

Total distance

[tex]{:}\longrightarrow[/tex][tex]\sf 100000+560=1000560cm [/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf 1000.56m[/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf 1.560km [/tex]

Answer:

Force constant of the spring (k) = 24.07 N/m

Concept/Theory:

The period [tex] \sf (T_s) [/tex] of a spring-mass system is proportional to the square root of the mass (m) and inversely proportional to the square root of the force constant of the spring (k).

Equation of period:

[tex] \boxed{ \bf{T_s = 2 \pi \sqrt{\dfrac{m}{k}}}}[/tex]

Explanation:

Mass = 0.250 kg

Period = 0.640 s

By substituting values in the equation, we get:

[tex] \rm \longrightarrow 0.640 = 2 \pi \sqrt{\dfrac{0.250}{k}} \\ \\ \rm \longrightarrow 2 \times 3.14 \sqrt{ \dfrac{0.25 0}{k} } = 0.640 \\ \\ \rm \longrightarrow 6.28 \sqrt{ \dfrac{0.250}{k} } = 0.640 \\ \\ \rm \longrightarrow \sqrt{ \dfrac{0.250}{k} } = \frac{0.640}{6.28} \\ \\ \rm \longrightarrow \frac{0.250}{k} = { \bigg(\frac{0.640}{6.28} \bigg) }^{2} \\ \\ \rm \longrightarrow \frac{k}{0.250} = \bigg( { \frac{6.28}{0.640} \bigg) }^{2} \\ \\ \rm \longrightarrow k = \bigg( { \frac{6.28}{0.640} \bigg) }^{2} \times 0.250 \\ \\ \rm \longrightarrow k = 24.07 \: N/m[/tex]

The force constant of the spring is approximately 24.038 newtons per meter.

As we are talking about Simple Harmonic Motion. In this exercise we need to determine the Spring Constant ([tex]k[/tex]), in newtons per meter, from the equation of the Period ([tex]T[/tex]), in seconds, which is described below:

[tex]T = 2\pi\cdot \sqrt{\frac{m}{k} }[/tex] (1)

Where [tex]m[/tex] is the mass of the moving element, in kilograms.

If we know that [tex]T = 0.640\,s[/tex] and [tex]m = 0.250\,kg[/tex], then the spring constant of the spring is:

[tex]0.640 = 2\pi\cdot \sqrt{\frac{0.250}{k} }[/tex]

[tex]\sqrt{\frac{0.250}{k} } \approx 0.102[/tex]

[tex]\frac{0.250}{k} \approx 0.0104[/tex]

[tex]k \approx 24.038\,\frac{N}{m}[/tex]

The force constant of the spring is approximately 24.038 newtons per meter.

Please see this question related to Simple Harmonic Motion for further details: https://brainly.com/question/17315536

One of the main uses of this device is to examine a beam of charged particles

The cathode ray tube is a device used to determine the charge flowing in a gas. When an electric field is set up with the help of metal electrodes, the cathode ray tends to bend towards the positive electrode.

Since the cathode ray bends towards the electrodes, it implies that it has a charge and the electrodes present help us determine the charge of the beam of charged particles and thus examine the beam of charged particles.

So, one of the main uses of this device is to examine a beam of charged particles.

Learn more about cathode ray tube here:

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Answer:

B

Explanation:

Answer:

15.02 m/s.

Explanation:

Given that the height of the hill, h= 11.5 m.

Combined mass, m= 54.8 kg

The initial velocity of the combined mass, u=0

Acceleration due to gravity, [tex]g = 9.81 m/s^2[/tex].

Angle of the path the horizontal, [tex]\theta = 36[/tex] degree.

Let A be the initial position and B be the final position of the sled as shown in the figure.

The path is frictionless so the drag force =0

The gravitational force acting on the combined mass in the downward direction, [tex]F= mg\cdots(i)[/tex]

The component of force acting in the direction of motion = [tex]F\sin \theta.[/tex]

Let [tex]a[/tex] be the acceleration of the combined mass, m, So,

[tex]F\sin \theta= ma[/tex]

[tex]\Rightarrow mg \sin \theta= ma[/tex] [ from equation (i)]

[tex]\Rightarrow a = g \sin \theta \cdots(ii).[/tex]

Let v be the final velocity of the combined mass.

Now, by using the equation of motion,

[tex]v^2=u^2+2as\\\\\Rightarrow v^2=0^2+2as\\\\ \Rightarrow v^2=2as\cdots(iii)[/tex]

Here, s is the displacement in the direction of motion,

So, s= AB

Now, in the right-angled triangle ABO,

[tex]\sin\theta = OA/AB= h/AB\\\\\Rightarrow AB = h/ \sin\theta\\\\\Rightarrow s = h/ \sin\theta\cdots(iv)[/tex]

Now,  from equations (ii), (iii) and (iv), we have

[tex]v^2= 2\times g \sin \theta \times \frac {h}{\sin\theta}\\\\\Rightarrow v^2= 2gh\\\\\Rightarrow v= \sqrt{2gh}[/tex]

By using the given values, we have

[tex]v= \sqrt{2\times 9.81\times 11.5}=\sqrt {225.63}\\\\\Rightarrow v = 15.02 m/s[/tex]

Hence, the speed of the combined mass at the bottom = 15.02 m/s.

The speed  of the sled at the bottom of the hill is [tex]15.02m/s[/tex]

The speed of sled is calculated by using Newton's law of motion,

                   [tex]v^{2} =u^{2} +2gh[/tex]

where u is initial velocity, v is final velocity , g is acceleration due to gravity and h is height.

Given that, [tex]u = 0, g = 9.81m/s^{2}[/tex] and [tex]h = 11.5 m[/tex]

Substitute values in above equation.

     [tex]v^{2}=0^{2}+2*9.81*11.5\\\\v^{2}=225.63\\\\v=\sqrt{225.63}=15.02m/s[/tex]

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As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.

Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.

The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:

n + (-w) = 0

n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N

The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be

f = 0.67 (51.94 N) ≈ 35 N

so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.

The block starts moving as soon as x = 35 N, at which point f = 35 N.

At any point after the block starts moving, we have

f = 0.48 (51.94 N) ≈ 25 N

so that x = 25 N is the required force to keep the block moving at a constant speed.

As x  is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force

Given data :

mass of block at rest ( m ) = 5.3 kg

Coefficient of static friction ( μ_s ) =0.67

Coefficient of kinetic friction is ( μ_k ) = 0.48

Horizontal force applied to block = x  

First step : magnitude of normal force ( n ) when object is at rest

n = w            where w = m*g

n - w = 0

n - ( 5.3 * 9.81 ) = 0     ∴  n = 51.94 N

Second step : Required magnitude of x before the movement of object

F =  μ_s * n

F = 0.67 * 51.94  = 34.79 N  ≈ 35 N

∴ The object will start moving once F and x = 35 N

Final step : Magnitude of x  after object start moving

F = μ_k  * n

  = 0.48 * 51.94 = 24.93 N  ≈ 25 N

∴ object will continue to accelerate at a constant speed once F and x = 25N

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