A box that is sliding across the floor experiences a net force of 10.0 N. If the box has a mass of 1.50 kg, what is the resulting acceleration of the box g

Answers

Answer 1

Answer:

a = 6.67 m/s²

Explanation:

F = 10.0 N

m = 1.50 kg

a = ?

F = ma

10.0 = (1.50)a

6.67 = a


Related Questions

find out the odd one and give reason (length, volume, time, mass​

Answers

Answer:

Time

Explanation:

The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.

A particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)​

Answers

Answer:

K = 0 J

Explanation:

Given that,

The mass of the particle, m = 1.2 mg

The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]

We need to find the kinetic energy of the particle at time t = 0 s.

At t = 0 s, the particle is at rest, v = 0

So,

[tex]K=\dfrac{1}{2}mv^2[/tex]

If v = 0,

[tex]K=0\ J[/tex]

So, the kinetic energy of the particle at time t = 0 s is 0 J.

A simple pendulum takes 2.00 s to make one compete swing. If we now triple the length, how long will it take for one complete swing

Answers

Answer:

3.464 seconds.

Explanation:

We know that we can write the period (the time for a complete swing) of a pendulum as:

[tex]T = 2*\pi*\sqrt{\frac{L}{g} }[/tex]

Where:

[tex]\pi = 3.14[/tex]

L is the length of the pendulum

g is the gravitational acceleration:

g = 9.8m/s^2

We know that the original period is of 2.00 s, then:

T = 2.00s

We can solve that for L, the original length:

[tex]2.00s = 2*3.14*\sqrt{\frac{L}{9.8m/s^2} }\\\\\frac{2s}{2*3.14} = \sqrt{\frac{L}{9.8m/s^2}}\\\\(\frac{2s}{2*3.14})^2*9.8m/s^2 = L = 0.994m[/tex]

So if we triple the length of the pendulum, we will have:

L' = 3*0.994m = 2.982m

The new period will be:

[tex]T = 2*3.14*\sqrt{\frac{2.982m}{9.8 m/s^2} } = 3.464s[/tex]

The new period will be 3.464 seconds.

If a car generates 22 hp when traveling at a steady 100 km/h , what must be the average force exerted on the car due to friction and air resistance

Answers

Answer:

The average force exerted on the car is 590.12 N.

Explanation:

Given that,

The power generated, P = 22 hp = 16405.4 W

Speed of the car, v = 100 km/h = 27.8 m/s

We need to find the average force exerted on the car due to friction and air resistance.

We know that,

Power, P = F v

Where

F is force exerted on the car

[tex]F=\dfrac{P}{v}\\\\F=\dfrac{16405.4}{27.8}\\\\F=590.12\ N[/tex]

So, the average force exerted on the car is 590.12 N.

Help me plssssssss cause I’m struggling

Answers

Answer:

I am pretty sure it is C

Explanation:

It can be found all over the universe

I think it’s c but I am not sure

Una cuerda horizontal tiene una longitud de 5 m y masa de 0,00145 kg. Si sobre esta cuerda se da un pulso generando una longitud de onda de 0,6 m y una frecuencia de 120 Hz. La tensión a la cual está sometida la cuerda es:

a. 1,5 N

b. 15,0 N

c. 3,1 N

d. 5,2 N

Answers

Answer:

Option (A) is correct.

Explanation:

A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is

mass of string, m = 0.00145 kg

Frequency, f = 120 Hz

wavelength = 0.6 m

Speed = frequency x wavelength

speed = 120 x 0.6 = 72 m/s

Let the tension is T.

Use the formula

[tex]v =\sqrt\frac{T L}{m}\\\\72 = \sqrt\frac{T\times 5}{0.00145}\\\\T = 1.5 N[/tex]

Option (A) is correct.

Question 3 of 10
Which statement describes the law of conservation of energy?
A. Air resistance has no effect on the energy of a system.
B. Energy cannot be created or destroyed.
C. The total energy in a system can only increase.
D. Energy cannot change forms.
هما
SUBMIT

Answers

Answer:

B . energy cannot be created or destroyed

Explain whether the unit of work is a fundamental or derived unit

Answers

Answer:

answer here

Explanation:

the unit of work is fundamental unit because it doesn't depend on other units.

__________________

Thx

Answer:

The SI unit of work is joule (J)

Explanation:

Joule is a derived unit. ∴ unit of work is a derived unit

1. An excited lithium atom emits a red light with wavelength a = 671nm. What is the corresponding photon energy? hc (6.63 x 10-34).S)(3.0 x 108m/s)​

Answers

Answer:

 E = 2,964 10⁻¹⁹ J

Explanation:

The energy of the photons is given by the Planck relation

          E = h f

the speed of light is related to wavelength and frequency

          c = λ f

we substitute

          E = h c /λ

let's reduce the magnitude to the SI system

          λ = 671 nm = 671 10⁻⁹ m

let's calculate

          E = 6.63 10⁻³⁴ 3 10⁸ /671 10⁻⁹

          E = 2,964 10⁻¹⁹ J

Which one of the following is not an example of convection? An eagle soars on an updraft of wind. A person gets a suntan on a beach. An electric heater warms a room. Smoke rises above a fire. Spaghetti is cooked in water.

Answers

Answer: The statement that is not an example of convection is (A person gets a suntan on a beach).

Explanation:

There are different modes of heat energy transfer which includes:

--> conduction

--> Radiation and

--> Convection

CONVECTION is a process by which heat energy is transferred in a fluid or air by the actual movement of the heated molecules. The cooler portion of the air surrounding a warmer part exerts a buoyant force on it. As the warmer part of the air moves, it is replaced by cooler air that is subsequently warmed.

Convection in gases is very common and gas expands more than liquid when subjected to high temperature.

--> it is used in bringing about the circulation of fresh air in the room in a process known as ventilation.Here, cool air is constantly being replaced with denser air ( warm air).

-->An electric heater warms a room and Smoke rises above a fire are typical example of convection in gases.

-->Spaghetti is cooked in water: As the water close to the burner warms, it rises to the top and boils. At the same time, cooler water on top moves downward to replace the rising hot water.

--> also the eagle uses convection current to stay afloat in the sky without flapping its wings to conserve energy.

But the option (A person gets a suntan on a beach) is an example of heat transfer through radiation. This is because the sun emits it's rays from the sky down to earth without any material medium unlike others. Therefore, this option is the ODD one out.

A 0.160kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.820m/s . It has a head-on collision with a 0.300kg glider that is moving to the left with a speed of 2.27m/s . Suppose the collision is elastic.
Part A
Find the magnitude of the final velocity of the 0.160kg glider. m/s
Part B
Find the direction of the final velocity of the 0.160kg glider.
i. to the right
ii. to the left
Part C
Find the magnitude of the final velocity of the 0.300kg glider. m/s
Part D
Find the direction of the final velocity of the 0.300kg glider.

Answers

Answer:

A) v_{f1} = -3.2 m / s,  B) LEFT , C) v_{f2} = -0.12 m / s,  D) LEFT

Explanation:

This is a collision exercise that can be solved using momentum conservation, for this we define a system formed by gliders, so that the forces during the collision are internal and the moment is conserved.

Let's use the subscript 1 for the lightest glider m1 = 0.160 kg and vo1 = 0.820 m / s

subscript 2 for the heaviest glider me² = 0.820 kg and vo2 = -2.27 m / s

Initial instant. Before the crash

          p₀ = m₁ v₀₁ + m₂ v₀₂

Final moment. After the crash

          p_f = m₁ v_{f1} + m₂ v_{f2}

          p₀ = p_f

          m₁ v₀₁ + m₂ v₀₂ = m₁ v_{f1} + m₂ v_{f2}

 as the shock is elastic, energy is conserved

         K₀ = K_f

         ½ m₁ v₀₁² + ½ m₂ v₀₂² = ½ m₁ [tex]v_{f1}^2[/tex] + ½ m₂ [tex]v_{f2}^2[/tex]

         m₁ (v₀₁² - v_{f1}²) = m₂ (v_{f2}² -v₀₂²)

let's make the relationship

         (a + b) (a-b) = a² -b²

         m₁ (v₀₁ + v_{f1}) (v₀₁-v+{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

let's write our two equations

         m₁ (v₀₁ -v_{f1}) = m₂ (v_(f2) - v₀₂)                                  (1)

         m₁ (v₀₁ + v_{f1}) (v₀₁-v_{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

we solve

         v₀₁ + v_{f2} = v_{f2} + v₀₂

we substitute in equation 1 and obtain

         M = m₁ + m₂

         [tex]v_{f1} = \frac{m_1-m_2}{M} v_o_1 + 2 \frac{m_2}{M} v_f_2[/tex]

         [tex]v_f_2 = \frac{2m_1}{M} v_o_1 + \frac{m_2-m_1}{M} v_o_2[/tex]vf2 = 2m1 / mm vo1 + m2-m1 / mm vo2

we calculate the values

         m₁ + m₂ = 0.160 +0.3000 = 0.46 kg

         v_{f1} = [tex]\frac{ 0.160 -0.300} {0.460} \ 0.820 + \frac{2 \ 0300}{0.460} \ (-2.27)[/tex]

         v_{f1} = -0,250 - 2,961

          v_{f1} = - 3,211 m / s

 

         v_{f2} = [tex]\frac{2 \ 0.160}{0.460} \ 0.820 + \frac{0.300 - 0.160}{0.460 } \ (-2.27)[/tex]

         v_{f2} = 0.570 - 0.6909

         v_{f2} = -0.12 m / s

now we can answer the different questions

A) v_{f1} = -3.2 m / s

B) the negative sign indicates that it moves to the left

C) v_{f2} = -0.12 m / s

D) the negative sign indicates that it moves to the LEFT

If a bale of hay behind the target exerts a constant friction force, how much farther will your arrow burry itself into the hay than the arrow from the younger shooter

Answers

Answer:

The arrow will bury itself farther by 3S₁

Explanation:

lets assume; the Arrow shot by me has a speed twice the speed of the arrow fired by the younger shooter

Given that ; acceleration is constant , Frictional force is constant

                    A₂ =   A₁

Vf²₂ - Vi²₂ / 2s₂  = Vf₁² - Vi₁² / 2s₁ ---- ( 1 )

final velocities = 0

Initial velocities : Vi₂ = 2(Vi₁ )

Back to equation 1

0 - (2Vi₁ )² / 2s₂ =  0 - Vi₁² / 2s₁

hence :

s₂ = 4s₁

hence the Arrow shot by me will burry itself farther by :

s₂ - s₁ = 3s₁

Note :  S1 = distance travelled by the arrow shot by the younger shooter

Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars increase their speed by 2.76 m/s, they then have the same kinetic energy. Calculate the original speeds of the two cars.

Answers

Let m be the mass of the second car, so the first car's mass is 2m.

Let K be the kinetic energy of the second car, so the first car's kinetic energy would be K/2.

Let u and v be the speeds of the first car and the second car, respectively. At the start,

• the first car has kinetic energy

K/2 = 1/2 (2m) u ² = mu ²   ==>   K = 2mu ²

• the second car starts with kinetic energy

K = 1/2 mv ²

It follows that

2mu ² = 1/2 mv ²

==>   4u ² = v ²

When their speeds are both increased by 2.76 m/s,

• the first car now has kinetic energy

1/2 (2m) (u + 2.76 m/s)² = m (u + 2.76 m/s)²

• the second car now has kinetic energy

1/2 m (v + 2.76 m/s)²

These two kinetic energies are equal, so

m (u + 2.76 m/s)² = 1/2 m (v + 2.76 m/s)²

==>   2 (u + 2.76 m/s)² = (v + 2.76 m/s)²

Solving the equations in bold gives u ≈ 1.95 m/s and v ≈ 3.90 m/s.

If one lawn mower causes an 80-dB sound level at a point nearby, four lawnmowers together would cause a sound level of ____________ at that point. a.92 dB b.84 dB c.86 dB d.none of the above

Answers

Answer:

The intensity of 4 lawn movers is 86 dB.

Explanation:

Intensity of one lawnmower = 80 dB

Let the intensity is I.

Use the formula of intensity

[tex]dB = 10 log\left ( \frac{I}{Io} \right )\\\\80=10log\left ( \frac{I}{Io} \right )\\\\10^8 = \frac{I}{10^{-12}}\\\\I = 10^{-4} W/m^2[/tex]

Now the intensity of 4 lawn movers is

[tex]dB = 10 log\left ( \frac{4I}{Io} \right )\\\\dB=10log\left ( \frac{4\times10^{-4}}{10^{-12}} \right )\\\\dB = 86 dB\\[/tex]

Is the actual height the puck reached greater or less than your prediction? Offer a possible reason why this might be.

Answers

Answer:

Answer to the following question is as follows;

Explanation:

The puck's real altitude is lower than ones projection. That's because the mechanism may not be completely frictionless. Electricity is nevertheless wasted owing to particle interactions such as friction, which might explain why the present the results is lower than predicted.

A typical incandescent light bulb consumes 75 W of power and has a mass of 20 g. You want to save electrical energy by dropping the bulb from a height great enough so that the kinetic energy of the bulb when it reaches the floor will be the same as the energy it took to keep the bulb on for 2.0 hours. From what height should you drop the bulb, assuming no air resistance and constant g?

Answers

Answer:

h = 2755102 m = 2755.102 km

Explanation:

According to the given condition:

Potential Energy = Energy Consumed by Bulb

[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]

where,

h = height = ?

P = Power of bulb = 75 W

t = time = (2 h)(3600 s/1 h) = 7200 s

m = mass of bulb = 20 g = 0.02 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]

h = 2755102 m = 2755.102 km

A 0.033-kg bullet is fired vertically at 222 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball

Answers

Answer:

The maximum height risen by the bullet-baseball system after the collision is 81.76 m.

Explanation:

Given;

mass of the bullet, m₁ = 0.033 kg

mass of the baseball, m₂ = 0.15 kg

initial velocity of the bullet, u₁ = 222 m/s

initial velocity of the baseball, u₂ = 0

let the common final velocity of the system after collision = v

Apply the principle of conservation of linear momentum to determine the common final velocity.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

0.033 x 222   +  0.15 x 0     = v(0.033 + 0.15)

7.326  =  v(0.183)

v = 7.326 / 0.183

v = 40.03 m/s

Let the height risen by the system after collision = h

Initial velocity of the system after collision = Vi = 40.03 m/s

At maximum height, the final velocity, Vf = 0

acceleration due to gravity for upward motion, g = -9.8 m/s²

[tex]v_f^2 = v_i^2 +2gh\\\\0 = 40.03^2 - (2\times 9.8)h\\\\19.6h = 1602.4\\\\h = \frac{1602.4}{19.6} \\\\h = 81.76 \ m[/tex]

Therefore, the maximum height risen by the bullet-baseball system after the collision is 81.76 m.

a student weighs 1200N they are standing in an elevator that is moving downwards at a constant speed of

Answers

Answer:

Elevator That Is Moving Downwards At A Constant Speed Of 4.9 M/S. What Is The Magnitude Of The Net Force Acing On The Student?

This problem has been solved!

This problem has been solved!See the answer

This problem has been solved!See the answerA student weighs 1200N. They are standing in an elevator that is moving downwards at a constant speed of 4.9 m/s. What is the magnitude of the net force acing on the student?

Explanation:

use this R= m(g-a), where R = reaction = weight, m= mass, a= acceleration and g= acceleration due to gravity

which is the correct formula for calculating the age of meteor right if using half life

Answers

Answer:

n × t_1/2

Exmplanation:

The age of meteorite is calculated by multiplying it's quantity n with the half life . This means that the formula is for age of this meteorite is;

Age of meteorite= n × t_1/2

where;

n = quantity of the meteorite

t_/2 = half life of the meteorite

Thus:

The correct formula is; n × t_1/2

Find the coefficient of kinetic friction between a 3.80-kg block and the horizontal surface on which it rests if an 87.0-N/m spring must be stretched by 6.50 cm to pull it with constant speed. Assume that the spring pulls in the horizontal direction.

Answers

Answer:

 μ = 0.15

Explanation:

Let's start by using Hooke's law to find the force applied to the block

          F = k x

          F = 87.0 0.065

          F = 5.655 N

Now we use the translational equilibrium relation since the block has no acceleration

          ∑ F = 0

          F -fr = 0

          F = fr

           

the expression for the friction force is

          fr = μ N

if we write Newton's second law for the y-axis

          N -W = 0

          N = W = mg

we substitute

          F = μ mg

          μ = F / mg

          μ = [tex]\frac{5.655}{3.8 \ 9.8}[/tex]

          μ = 0.15

An electric eel can generate a 180-V, 0.1-A shock for stunning its prey. What is the eel's power output

Answers

Power output = volts x amps

Power output = 170 volts x 0.1 amps

Power output = 18 watts

2. How do the phytochemicals present in various foods help us?

Answers

Answer:

Phytochemicals are compounds that are produced by plants ("phyto" means "plant"). They are found in fruits, vegetables, grains, beans, and other plants. Some of these phytochemicals are believed to protect cells from damage that could lead to cancer.

A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.0 in. (50.8 mm) is pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (f).
a. Plot the data as engineering stress versus engineering strain.
b. Compute the modulus of elasticity.
c. Determine the yield strength at a strain offset of 0.002.
d. Determine the tensile strength of this alloy.
e. What is the approximate ductility, in percent elongation?
f. Compute the strain energy density up to yielding (modulus of resilience).
( Load in N Load in lb Length in mm Length in in. 2.000 2.002 2.004 2.006 2.008 2.010 2.020 2.040 2.080 2.120 2.160 2.200 2.240 2.270 2.300 2.330 Fracture 50.800 7330 15,100 3400 23,100 5200 30,400 6850 34,400 7750 38,400 8650 41,3009300 44,800 10,100 46,200 10,400 53, 47,300 10,650 54.864 47,500 10,700 55.880 46,100 10,400 44,800 10,100 42,600 9600 3,400 8200 Fracture Fracture Fracture 50.851 50.902 50.952 51.003 51.054 1650 51.308 51.816 52.832 848 56.896 57.658 58.420 59.182

Answers

Answer:

A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.0 in. (50.8 mm) is pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (f).

The force an ideal spring exerts on an object is given by , where measures the displacement of the object from its equilibrium position. If , how much work is done by this force as the object moves from to

Answers

Answer:

The correct answer is "1.2 J".

Explanation:

Seems that the given question is incomplete. Find the attachment of the complete query.

According to the question,

x₁ = -0.20 mx₂ = 0 mk = 60 N/m

Now,

⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]

⇒      [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]

⇒      [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]

By putting the values, we get

⇒      [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]

⇒      [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]

⇒      [tex]=1.2 \ J[/tex]

SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.7 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building

Answers

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

a vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?​

Answers

Answer:

Explanation:

Speed= distance/time

Or time = distance/speed

According to your question

Speed=15m/s

and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s

D= 1.2km = 1.2×1000m =1200meter

Time = distance/ speed

1200/15 =80second

Or. 1min and 20 sec will be your answer.

two point charges with charge q are initially separated by a distance d. if you double the charge on both charges, how far should the charges be separated for the potential energy between them to remain the same

Answers

Answer:

  r ’= 4 r

Explanation:

Electric potential energy is

          U = [tex]k \frac{q_1q_2}{r_{12}}[/tex]k q1q2 / r12

in this exercise

          q₁ = q₂ = q

          U = k q² / r

for when the charge change

           U ’= k q’² / r’

indicate that

      q ’= 2q

      U ’= U

we substitute

           U = k (2q) ² / r ’

           U = 4 k q² / r ’

we substitute

           [tex]k \ \frac{q^2}{r} = 4 k \ \frac{q^2}{r'}[/tex]k q² / r = 4 k q² / r ’

           r ’= 4 r

) Efficiency of a lever is always less than hundred percent.​

Answers

Yes. Because it opposes the law of friction

I hope this helps.

Explanation:

Please mark me brainliest

If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact would it make? And could you also show me the equation to solve similar problems myself. Thank you.

Answers

Answer:

The impact force is 98000 N.

Explanation:

mass = 10 tons

The impact force is the weight of the object.

Weight =mass x gravity

W = 10 x 1000 x 9.8

W = 98000 N

The impact force is 98000 N.

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Newtons. What is the frequency of this mode of vibration

Answers

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

[tex]F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz[/tex]

Therefore, the frequency of this mode of vibration is 138.87 Hz

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