please help im failing!!!

Answer:

108 N

Explanation:

Use Newton's second law.

F = ma

F = (90 kg) (1.2 m/s²)

F = 108 N

Answer:

a) t = 6.62 s

b) x = 238.6 m

c) H = 53.7 m

Explanation:

a) We can find the time of flight as follows:

[tex] y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} [/tex]

Where:

[tex]y_{f}[/tex] is the final height = 0

[tex]y_{0}[/tex] is the initial height = 0

[tex]v_{0_{y}}[/tex] is the initial vertical velocity of the stone

t: is the time

g: is the gravity = 9.81 m/s²

[tex] v_{0}sin(42)t - \frac{1}{2}gt^{2} = 0 [/tex]      

[tex] 48.5 m/s*sin(42)*t - \frac{1}{2}9.81 m/s^{2}*t^{2} = 0 [/tex]  

By solving the above quadratic equation we have:  

t = 6.62 s

b) The maximum range is:

[tex] x = v_{0_{x}}t = 48.5 m/s*cos(42)*6.62 s = 238.6 m [/tex]

c) The maximum height (H) can be found knowing that at this height the final vertical velocity of the stone is zero:

[tex] v_{f_{y}}^{2} = v_{0_{y}}^{2} - 2gH [/tex]  

[tex] H = \frac{v_{0_{y}}^{2} - v_{f_{y}}^{2}}{2g} = \frac{(48.5 m/s*sin(42))^{2} - 0}{2*9.81 m/s^{2}} = 53.7 m [/tex]

I hope it helps you!

An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.

Since the object must be moved away to a distance greater than the radius of the Earth, then change in gravitational potential energy must be based on Newton's Law of Gravitation.

By the Work-Energy Theorem, the work ([tex]W[/tex]), in joules, done on the object is equal to the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:

[tex]W = U_{g}[/tex] (1)

[tex]W = -G\cdot m\cdot M\cdot \left(\frac{1}{r_{f}}-\frac{1}{r_{o}} \right)[/tex] (1b)

Where:

If we know that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]m = 1000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]r_{o} = 6.371\times 10^{6}\,m[/tex] and [tex]r_{f} = 19.113\times 10^{6}\,m[/tex], then the energy required to move the object from the Earth's surface is:

[tex]W = -\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (1000\,kg)\cdot (5.972\times 10^{24}\,kg)\cdot \left[\frac{1}{19.113\times 10^{6}\,m} - \frac{1}{6.371\times 10^{6}\,m} \right][/tex][tex]W = 4.171\times 10^{10}\,J[/tex]

An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.

We kindly invite to check this question on gravitational potential energy: https://brainly.com/question/19768887

Answer:

The speed the apple must be thrown in order for it to land in the basket is  2.554 m/s.

Explanation:

Given;

height above the ground, h = 3.0 m

horizontal distance, X = 2.0 m

The time to drop from the given height;

h = ¹/₂gt²

[tex]t = \sqrt{\frac{2h}{g} }\\\\t = \sqrt{\frac{2*3}{9.8} }\\\\t = 0.783 \ s[/tex]

The horizontal speed traveled by the apple is given by;

vₓ = X / t

vₓ = 2 / 0.783

vₓ = 2.554 m/s

Therefore, the speed the apple must be thrown in order for it to land in the basket is  2.554 m/s.

Answer: length,  luminous intensity,mass, time, temperature, electric current, amount of a substance.

Explanation:

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{25}{2} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

a) the centripetal force causing the cat to turn in a circular path is due to friction between the car's tyres and the surface of the road.

b)

1) a larger centripetal force is required if the car travels faster

2) a larger centripetal force is needed when the bend is less curved.

3) when a car has more passengers, the mass of the car increases, therefore, a larger force is required to move the car.

Explanation:

The friction between the car's tire and the surface provides centripetal force and if the car travels faster, the bend is curved, or the car has more passengers it requires a larger centripetal force.

When an object is moving in a curve path then a force is acting radically in the object this force is called centripetal force, this force helps to maintain the object in the curve e.g. if a car is moving in a well, so the centripetal force is responsible, so the car is not falling. The SI unit used to represent centripetal force is Newton.

A. The friction between the car's tire and the curve surface is responsible for providing the centripetal force, and it also depends on the car's speed.

B.  If a car is traveling faster than a larger centripetal force is required to maintain the curve path of the car, means if the friction between the tire and surface decreases the centripetal force also decreases.

When the bend is less curved, the radius of the curve increase and thus require more centripetal force.

When people in the car increase then the mass of the car also increases so due to this the gravitation force is now more on the car so more centripetal force I required to maintain the curve path of the car.

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Answer:C,(Calcium,an alkaline earth metal with an oxidation number of +2 will form an ionic bond with chlorine,a halogen in group VllA with an oxidation number of -1 called calcium chloride (CaCl2)

Explanation:

on USAtestprep !!

Calcium, an alkaline earth metal with an oxidation number of +2 will form

an ionic bond with chlorine, a halogen in group VIIA with an oxidation

number of -1 called calcium chloride (CaCl₂). This is correct statement.

Simply said, the number assigned to each element in a chemical combination is the definition of an oxidation number. The total number of electrons that an atom in a molecule can share, lose, or gain while forming a chemical bond with an atom of a different element is known as the oxidation number.

Also known as oxidation state, oxidation number is a numerical value. But depending on whether we take into account the atoms' electronegativity or not, these phrases might occasionally have a different meaning. In coordination chemistry, the term "oxidation number" is often used.

According to Periodic table:  calcium is a alkaline earth metal with an oxidation number of +2 whereas chlorine is a halogen in group VIIA with an oxidation number of -1. When they reacts chemically, they form an ionic compound named calcium chloride having chemical formula CaCl₂.

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Answer:

If a cow has a greater mass than the tiger, it has greater momentum, but only if they are moving at the same speed, or the tiger is going slower (i think)

Explanation:

the quantity of motion of a moving body, measured as a product of its mass and velocity.

Answer:

The acceleration of the racecar is [tex]\mathbf{11.17~m/s^2}[/tex]

Explanation:

Uniformly Accelerated Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Following the definition above, the acceleration is defined as:

[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]

Where a is the constant acceleration, vo the initial speed, vf the final speed, and t the time.

The racecar goes from vo=18.5 m/s to vf=46.1 m/s in t=2.47 seconds, thus the acceleration is:

[tex]\displaystyle a=\frac{46.1-18.5}{2.47}[/tex]

[tex]\displaystyle a=\frac{27.6}{2.47}[/tex]

[tex]a = 11.17~m/s^2[/tex]

The acceleration of the racecar is [tex]\mathbf{11.17~m/s^2}[/tex]

Answer:

Work Function = 3.53 x 10⁻¹⁹ J = 2.2 eV

Explanation:

The work function of the metal metal can be found as follows:

Energy of Photon = Work Function + K.E

hc/λ = Work Function + K.E

Work Function = hc/λ - K.E

where,

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = speed of light  = 3 x 10⁸ m/s

λ = wavelength of photons = 240 nm = 2.4 x 10⁻⁷ m

K.E = Maximum Kinetic Energy = (2.97 eV)(1.6 x 10⁻¹⁹ J/1 eV) =  4.752 x 10⁻¹⁹ J

Therefore,

Work Function = (6.625 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.4 x 10⁻⁷ m) - 4.752 x 10⁻¹⁹ J

Work Function = 8.281 x 10⁻¹⁹ J - 4.752 x 10⁻¹⁹ J

Work Function = 3.53 x 10⁻¹⁹ J = 2.2 eV

Answer:

The value is  [tex]\mu_k = 0.03[/tex]

Explanation:

From the question we are told that

  The force applied on the ice is [tex]F = 29.4 N[/tex]

   The mass of the ice block is  [tex]m = 100 \ kg[/tex]

Generally for the ice block to move at constant speed , the force applied on it must be equal to the kinetic frictional force which is mathematically represented as

         [tex]F_F = m* g * \mu_k[/tex]

=>      [tex]F = F_F = m* g * \mu_k[/tex]

=>      [tex]29.4 = 100 * 9.8 * \mu_k[/tex]

=>       [tex]\mu_k = 0.03[/tex]

Answer:

Its color

Explanation: I got it right

Answer:

knee extension is the muscle function that will be difficult to perform.

Explanation:

Barb was kicked in the anterior thigh. Now, the thigh muscles performs a combined operation of moving the knee and leg and they reside in the following compartments.

- Anterior compartment which is composed of knee joint extension and thigh flexion.

- Lateral Compartment which is composed of the tensor fasciae latae, which is a tiny muscle that abducts and centrally will make the thigh to rotate.

-Medial compartment which involves thigh addiction which is rotating of the thigh around the hips.

- Posterior compartment which involves knee joint flexion and high extension.

Thus, from the different compartments listed above, we can see that the muscles that extend the knee and flex the thigh all lie in the anterior compartment of the upper leg.

Thus, we can conclude that knee extension is the muscle function that will be difficult to perform.

This question is incomplete, the complete question is;

A coil consists of 200 turns of wire. Each turn is a square of side 18 cm, and a uniform magnetic field directed perpendicular to the plane of the coil is turned on. If the field changes linearly from 0 to 0.50 T in 0.80 s,

a) what is the magnitude of the induced emf in the coil while the field is changing?

b) if the resistance of the coil is 2.0, what is the magnitude of the induced current in the coil while the field is changing?

Answer:

a) the magnitude of the induced emf in the coil is 4.05 V

b) induced current in the coil I is 2.025 A

Explanation:

Given that;

side of turn a = 18 cm = 0.18 m

no. of turns N = 200

dB = 0.50 T

time t = 0.80 sec

(a)

what is the magnitude of the induced emf in the coil while the field is changing?

we know that the magnetic flux is equal to the product of the magnetic field in a loop and the area of the loop so;

 ∅ = NBA

expression for the electromotive force is expressed as;

∈ = d∅/dt

Now replace NBA for ∅ in the above equation.

∈ = d(NBA) / dt

= NA(dB/dt)

The expression for the area of each square turn is expressed as follows

A = a²

a is the side of the turn

so we substitute the value of a

A = (0.18) ²

A= 0.0324 m²

 As  earlier derived

formula for the electromotive force is as follows:

∈  = NA(dB/dt)

so we substitute all our values

∈ =  (200)(0.0324m²) (0.50T/0.80s)

∈ = 6.48 × 0.625

∈ = 4.05 V

Therefore the magnitude of the induced emf in the coil is 4.05 V

(b)

if the resistance of the coil is 2.0, what is the magnitude of the induced current in the coil while the field is changing?

we know that the current induced in the circuit is equal to the ratio between the electromotive forces to the resistance of the ring so;

I = ∈ / R

given that; resistance of the coil = 2.0

so we substitute

I = 4.05 / 2.0

I = 2.025 A

Therefore induced current in the coil I is 2.025 A

Answer:

Explanation:

charge on the shell = 4πR²  x charge density

= 4 x 3.14 x 7.5² x 10⁻⁴ x 10⁻³ C

= 706.5 x 10⁻⁷ C  

electric field = k Q / d² , d is distance of point from centre

= 9 x 10⁹ x 706.5 x 10⁻⁷ / 10²

= 6358.5 N /C

= 6.358 kV / m

Answer:

I found the experience tasking

Explanation:

I wouldn't say it was hard, neither was it easy. I'd rather go for something like it being tasking. It's worthy of note that it was my first time, and I think it's very normal especially when one hasn't been doing something of that nature previously. Of course I did my draft, which unsurprisingly happened to be not good enough, and I had to look for templates to guide me through the acceptable way.

I still did it in my own way, but in the right way. Ever since then though, I have never stuttered when writing application letters, as it had since then seem inborn

The experience of writing a bio-data or an application letter was quite a tasking goal, where proper structure is required.

The given problem is based on the fundamentals of bio-data. Biodata is a document that is used to display the biographical data about the work experience in any organization.

Thus, it is concluded that writing a bio-data or an application letter was quite a tasking goal, where proper structure is required.

Learn more about the biodata here:

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Explanation:

Step one:

given dara

mass of ball m=0.kg

initial velocity u= 10 m/s

final velocity v=10m/s

Required

momentum before and after impact

the expression for momentum P

P=mv

before impact

P=mv

P1=0.5*10

P1=5(kg⋅m/s)

after impact

P=mu

P2=0.5*10

P2=5(kg⋅m/s)

change  in momentum is =P1-P2= 5-5=0

yes, momentum is conserved.

60 m

Concept Used:

We know that the area under a velocity-time graph represents the Displacement of the body

Displacement in the Last 6 seconds:

To find the Displacement in the last 6 seconds, we will find the area under the graph between x = 4 and x = 10

We can see that the shape formed is a rectangle also shown in the given graph. So, the area of the rectangle is the Displacement of the car in the last 6 seconds

Area of the Rectangle:

From the graph, we know that the rectangle is 10 (m/s) tall and 6 (s) wide

Area of Rectangle=  length*Breadth        

replacing the values

Area = 10 (m/s) * 6 (s)

Area = 60 m

Hence, the car travelled 60 m in the last 6 seconds of the graph

Answer:

Electromagnetic waves

Electromagnetic waves bring energy into a system by virtue of their electric and magnetic fields. These fields can exert forces and move charges in the system and, thus, do work on them. However, there is energy in an electromagnetic wave itself, whether it is absorbed or not.

So the answer is B electrical energy

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Answer:

21.21 m/s

Explanation:

Let KE₁ represent the initial kinetic energy.

Let v₁ represent the initial velocity.

Let KE₂ represent the final kinetic energy.

Let v₂ represent the final velocity.

Next, the data obtained from the question:

Initial velocity (v₁) = 15 m/s

Initial kinetic Energy (KE₁) = E

Final final energy (KE₂) = double the initial kinetic energy = 2E

Final velocity (v₂) =?

Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:

KE = ½mv²

NOTE: Mass (m) = constant (since we are considering the same car)

KE₁/v₁² = KE₂/v₂²

E /15² = 2E/v₂²

E/225 = 2E/v₂²

Cross multiply

E × v₂² = 225 × 2E

E × v₂² = 450E

Divide both side by E

v₂² = 450E /E

v₂² = 450

Take the square root of both side.

v₂ = √450

v₂ = 21.21 m/s

Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.

For the car to be able to double its kinetic energy, it would need to travel at a speed of approximately 21.21m/s.

Given the data in the question;

Speed of the car; [tex]v_1 = 15m/s[/tex]

Speed for the car to double its kinetic energy; [tex]v_2 = \ ?[/tex]

Using the expression for kinetic energy:

[tex]K_E = \frac{1}{2}mv^2[/tex]

Where m is the mass and v is the velocity

Now, Initial Kinetic Energy will be;

[tex]K_E_1 = \frac{1}{2}mv_1^2 \\\\K_E_1 = \frac{1}{2}*m*(15)^2\\\\K_E_1 = \frac{1}{2}*m*225\\\\K_E_1 = 112.5*m[/tex]

For the kinetic energy to become double

[tex]K_E_2 = 2 * K_E_1\\\\\frac{1}{2}mv_2^2 = 2 * ( 112.5 * m)\\\\\frac{1}{2}mv_2^2 = 2m( 112.5)\\\\v_2^2 = 4( 112.5)\\\\v_2^2 = 450\\\\v_2 = \sqrt{450}\\\\v_2 = 21.21m/s[/tex]

Therefore, for the car to be able to double its kinetic energy, it would need to travel at a speed of approximately 21.21m/s.

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Answer:

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question we have

mass = 1.5 × 8

We have the final answer as

Hope this helps you

Answer: True

Explanation: Guessed on A p E x and it was correct

Answer:

True

Explanation:

a p e x

Answer:

The answer is "[tex](1.265 \pm 0.010) \ s \ and \ 0.709 \%[/tex]"

Explanation:

In point i:

[tex]T_{theo}= 2\pi \sqrt{\frac{l}{g}}[/tex]

        [tex]=2\pi\sqrt{\frac{0.397}{9.8}}\\\\= 1.265 \ s[/tex]

If  error in the theoretical time period :

[tex]\frac{\Delta T_{theo}}{T_theo} = \frac{1}{2} \frac{\Delta l }{l}\\\\\Delta T_{theo} = 1.265 \times \frac{1}{2} \times \frac{0.006}{0.397}[/tex]

           [tex]= 0.010 \ s[/tex]

 [tex]T_{theo} = (1.265 \pm 0.010) \ s[/tex]

In point ii:

[tex]\% \ difference = \frac{|T_{exp} -T_{theo}|}{\frac{T_{exp}+T_{theo}}{2}} \times 100[/tex]

Answer:

The acceleration of the particle as it travels one semicircle is

[tex]a= \frac{\pi R}{t^2}[/tex]

Explanation:'

Kindly see attached a sketch of a semi-circle

Step one:

given data

velocity =v

let the time taken be t

The path PQM is the distance covered

so distance [tex]d= \pi R[/tex]

we know that time= distance/velocity

t= πR/v

step two:

velocity =distance/time

[tex]velocity=\frac{\pi R}{t}[/tex]

also, we know that acceleration is velocity/time

[tex]a= \frac{\pi R}{\frac{t}{t} }[/tex]

[tex]a= \frac{\pi R}{t}*\frac{1}{t}[/tex]

[tex]a= \frac{\pi R}{t^2}[/tex]

Answer:

coral reef

Explanation:

Answer:

True

Explanation:

have a good day:)

Answer: This statement is True

Answer:

Pic not clear.........