A weight lifter is trying to lift a 1,500 N weight but can apply a force of only 1,200 N on the weight. One of his friends helps him lift it at a constant velocity. What force was applied to the rope

Answer:

See below

Explanation:

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( hope it helped <3 )

Answer:

1144.8

Explanation:

1 meter per second is equal to 3.6 kilometers per hour

Answer:

Explanation:

a. WEIGHT: As we have seen, weight is the gravitational force exerted on an object by the Earth (or any other celestial body). If an object is near the Earth's surface and has mass, then the object has a weight. The magnitude of its weight is w = mg and its direction is toward the center of the Earth.

Explanation:

Given that,

Initial velocity, u = 11.3 m/s

Angle above the horizontal, [tex]\theta=35^{\circ}[/tex]

Time of flight :

[tex]t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s[/tex]

Horizontal distance traveled  is given by :

x = ut

x = 11.3 m/s × 1.32 s

x = 14.916 m

Maximum height is given by :

[tex]H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m[/tex]

Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.

Explanation:

P = ρgh

P = (9.6×10² kg/m³) (10 N/kg) (0.16 m)

P = 1536 Pa

Rounded to two significant figures, the pressure is 1500 Pa, or 1.5 kPa.

Answer:

The range is 35.35 m

Explanation:

Projectile Motion

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and [tex]g=9.8m/s^2[/tex] the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

[tex]\displaystyle d={\frac {v_o^{2}\sin(2\theta )}{g}}[/tex]

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

[tex]\displaystyle d={\frac {20^{2}\sin(2\cdot 30^\circ )}{9.8}}[/tex]

[tex]\displaystyle d={\frac {400\sin(60^\circ )}{9.8}}[/tex]

[tex]d=35.35\ m[/tex]

The range is 35.35 m

Explanation:

Electronegativity is a measure of an atom's ability to attract shared electrons to itself. On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group.

Answer:

A)

Explanation:

I took the test and got it right. Flvs

Answer:B

Explanation:

Mass

More mass is added to a wheelbarrow, and a larger force is required to move it - illustrates Newton's second law.

An object's push or pull is seen as exerting a force. The interaction of the objects produces push and pull. You can also use words like stretch and squeeze to describe force.

The definition of force in physics is: The push or pull on a mass-containing item changes its velocity.

Newton’s second law of motion states that “Force is equal to the rate of change of momentum. For a constant mass, force equals mass times acceleration."

Hence more mass is added to a wheelbarrow, and a larger force is required to accelerate, i.e., move it.

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Answer:

Cation

Explanation:

Best way to remember this one is a CATion is PAWS-itive!

Answer:

C

Explanation:

Answer:

45 degrees

Explanation:

a projectile travels the farthest when it is launches at the angle of 45 degree.

Answer:

2.a centripetal force.

Answer:

W = 1493.9 J = 1.49 KJ

Explanation:

The work done by the elevator on the object will be equal to the gain in is potential energy:

W = ΔP.E

W = mgΔh

where,

W = Work = ?

m = mass of object = 7.4 kg

g = 9.8 m/s²

Δh = gain in height = 20.6 m

Therefore,

W = (7.4 kg)(9.8 m/s²)(20.6 m)

W = 1493.9 J = 1.49 KJ

Answer:

The car that is accelerating is B a car that rounds a curve at a constant speed

Explanation:

Although all of the cars are at a constant speed or not moving acceleration is the change in speed or the change of directions therefore making the only car changing directions your answer.

Answer:

its B (:

Explanation:

i  took the test :)

Answer:

a) El número de neutrones de Cl-35 es 18 y del Cl-37 es 20.

b) La masa atómica media del cloro es 35.5 uma.

c) En 15 g de cloro hay 2.55x10²³ átomos.    

Explanation:

a) Para calcular los protones y neutrones de los isótopos Cl-35 y Cl-37 debemos usar la siguiente fórmula:

[tex] A = Z + N [/tex]

En donde:

Z: es el número de protones = 17

N: es el número de neutrones

A: es la masa atómica

Los isótopos de un elemento dado tienen el mismo número de protones (número atómico) y en ellos sólo varia la la masa atómica (y por ende el número de neutrones); por lo que el Cl-35 y Cl-37 tienen igual Z (17).

Para el Cl-35 tenemos:

[tex] N = A - Z = 35 - 17 = 18 [/tex]

Y para el Cl-37 tenemos:

[tex]N = 37 - 17 = 20[/tex]

Por lo tanto, el número de neutrones de Cl-35 es 18 y del Cl-37 es 20.

b) La masa atómica media del cloro está dada por:

[tex]M_{Cl} = A_{Cl-35}*\%_{Cl-35} + A_{Cl-37}*\%_{Cl-37} = 35 u*\frac{75}{100} + 37 u*\frac{25}{100} = 35.5 u[/tex]  

Entonces, la masa atómica media del cloro es 35.5 uma.

c) Para encontrar el número de átomos de Cl que hay en 15 g debemos usar el número de Avogadro (6.022x10²³ átomos/mol):

[tex] N = \frac{1 mol}{35.5 g}*15 g*\frac{6.022 \cdot 10^{23} atomos}{1 mol} = 2.55 \cdot 10^{23} atomos [/tex]

Finalmente, en 15 g de cloro hay 2.55x10²³ átomos.

Espero que te sea de utilidad!          

Answer:

Explanation:

Gneiss is a high grade metamorphic rock, meaning that it has been subjected to higher temperatures and pressures than schist. However, unlike slate and schist, gneiss does not preferentially break along planes of foliation because less than 50% of the minerals formed during the metamorphism are aligned in thin layers.

Answer:

The arrow of the second archer standing on a high building will have more potential energy.

The arrow of the first archer standing on ground will have more kinetic energy.

Explanation:

POTENTIAL ENERGY:

The potential energy of an object depends upon its height, as given in the formula:

P.E = mgh

Hence, the arrow with the greater height will have more potential energy.

Therefore, the arrow of the second archer standing on a high building will have more potential energy.

KINETIC ENERGY:

The kinetic energy of an object depends upon its speed, as given in the formula:

K.E = (1/2)mv²

Hence, the arrow with the greater speed will have more kinetic energy energy. Since, the second arrow is stationary, it will have zero kinetic energy. But, the first arrow will have some K.E due to its speed.

Therefore, the arrow of the first archer standing on ground will have more kinetic energy.

(a)  The second archer standing at the top of the building will have more potential energy.

(b) The first archer on the ground level will have the more kinetic energy.

The potential energy of an object is the energy possessed by the object by virtue its position above above the ground.

P.E = mgh

where;

Thus, the second archer standing at the top of the building will have more potential energy.

(b) The kinetic energy of an object is the energy possessed by the object due to its motion.

K.E = ¹/₂mv²

where;

An object in motion, has zero velocity at maximum height but maximum velocity on the ground level.

Thus, the first archer on the ground level will have the more kinetic energy.

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Answer:

plane LA534 will have to decrease it speed by twice that of plane  LA639.

Explanation:

Since both planes want to land there must be a decrease in speed.

We are told that plane  LA534 velocity is twice that of plane LA639.

Now for the two planes to land without a crash. the opposite or reverse will be the case in terms of deceleration, that is

Plane LA639 will have to decrease its speed by twice the decrease in speed of plane  LA534

Answer:

12.18

Explanation:

The maximum height that the orange will reach is 12.18 m.

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion.

You throw an orange out a window at a height of 12.00 meters upwards at an angle of 32° to the horizontal at a velocity of 3.5 m/s.

If u is the initial speed, the maximum height of the ball from the point of throwing is

H = u² sin²θ / 2g

Put the values, we get

H = 3.5² sin²32° / (2x9.81)

H = 0.1753 m

So, the maximum height calculated from the ground is

12 + 0.18 = 12.18 m

Thus ,the maximum height that the orange will reach is 12.18 m.

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Answer:

There are little tiny batteries that are contained within the molecules of the food. The sun charges these batteries, and when the food is eaten, the energy from the batteries flows through the acids in the stomach, which gives the person or animal energy. In gasoline, the energy is created when the gasoline is set on fire and burned, there is no energy within it already. So they are different forms of energy. Hope this helps!

Answer:

Mass, m = 71.42 kg

Explanation:

Given that,

Normal force acting on a box, F = 700 N

We need to find the mass of a box. Let it is m. Normal force acting on an object is balanced by its weight such that,

F = mg

Where m is the mass of the box

[tex]m=\dfrac{F}{g}\\\\m=\dfrac{700\ N}{9.8\ m/s^2}\\\\m=71.42\ kg[/tex]

So, the mass of the box is 71.42 kg.

Answer:

1) The time it takes the diver to move off the end of the diving board to the pool surface, [tex]t_w[/tex], is approximately 1.392 seconds

2) The horizontal distance from the edge of the pool to where the diver enters the water, [tex]d_w[/tex], is approximately 5.76 meters

Explanation:

1) The given parameters are;

The height of the diving board above the pool's surface, h = 9.5 m

The length by which the diving board over hangs the pool L = 2 m

The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s

Taking [tex]t_w[/tex] = The time it takes the diver to move off the end of the diving board to the pool surface

Therefore, we have from the equation of free fall;

h = 1/2 × g × [tex]t_w[/tex]²

Where;

g = The acceleration due to gravity = 9.81 m/s²

Substituting the values, gives;

9.5 = 1/2 × 9.81 × [tex]t_w[/tex]²

[tex]t_w[/tex] = √(9.5/(1/2 × 9.81)) ≈ 1.392 s

The time it takes the diver to move off the end of the diving board to the pool surface = [tex]t_w[/tex] ≈ 1.392 s

2) The horizontal distance, [tex]d_w[/tex], in meters from the edge of the pool to where the diver enters the water is given as follows;

[tex]d_w[/tex] = L + v₀ × [tex]t_w[/tex] = 2 + 2.7× 1.392 ≈ 5.76 m

∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.

Answer:

The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].

Explanation:

We can estimate the absolute uncertainty by the definition of total differential. That is:

[tex]\Delta \rho \approx \frac{\partial \rho}{\partial m}\cdot \Delta m + \frac{\partial \rho}{\partial d}\cdot \Delta d + \frac{\partial \rho}{\partial l}\cdot \Delta l[/tex] (1)

Where:

[tex]\frac{\partial \rho}{\partial m}[/tex] - Partial derivative of the density with respect to mass, measured in [tex]\frac{1}{mm^{3}}[/tex].

[tex]\frac{\partial \rho}{\partial d}[/tex] - Partial derivative of the density with respect to diameter, measured in grams per cubic milimeter.

[tex]\frac{\partial \rho}{\partial l}[/tex] - Partial derivative of the density with respect to length, measured in grams per cubic milimeter.

[tex]\Delta m[/tex] - Mass uncertainty, measured in grams.

[tex]\Delta d[/tex] - Diameter uncertainty, measured in milimeters.

[tex]\Delta l[/tex] - Length uncertainty, measured in milimeters.

[tex]\Delta \rho[/tex] - Density uncertainty, measured in grams per cubic milimeters.

Partial derivatives are, respectively:

[tex]\frac{\partial \rho}{\partial m} = \frac{4}{\pi\cdot d^{2}\cdot l}[/tex] (2)

[tex]\frac{\partial \rho}{\partial d} = -\frac{8\cdot m}{\pi\cdot d^{3}\cdot l}[/tex] (3)

[tex]\frac{\partial \rho}{\partial l} = - \frac{4\cdot m}{\pi\cdot d^{2}\cdot l^{2}}[/tex] (4)

And we expand (1) as follows:

[tex]\Delta \rho \approx \frac{4\cdot \Delta m}{\pi\cdot d^{2}\cdot l} - \frac{8\cdot m\cdot \Delta d}{\pi\cdot d^{3}\cdot l}-\frac{4\cdot m\cdot \Delta l}{\pi\cdot d^{2}\cdot l^{2}}[/tex]

[tex]\Delta \rho \approx \left(\frac{4}{\pi\cdot d^{2}\cdot l}\right)\cdot \left(\Delta m -\frac{m\cdot \Delta d}{d}-\frac{m \cdot \Delta l}{l} \right)[/tex] (5)

If we know that [tex]d = 2\,mm[/tex], [tex]l = 25\,mm[/tex], [tex]m = 6.2\,g[/tex], [tex]\Delta m = \pm 0.1\,g[/tex], [tex]\Delta d = \pm 0.01\,mm[/tex] and [tex]\Delta l = \pm 0.1\,mm[/tex], then the absolute uncertainty is:

[tex]\Delta \rho \approx \pm\left[\frac{4}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)} \right]\cdot \left[(0.1\,g)-\frac{(6.2\,g)\cdot (0.01\,mm)}{2\,mm} -\frac{(6.2\,g)\cdot (0.1\,mm)}{25\,mm} \right][/tex]

[tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex]

And the expected density is:

[tex]\rho = \frac{4\cdot m}{\pi\cdot d^{2}\cdot l}[/tex] (6)

[tex]\rho = \frac{4\cdot (6.2\,g)}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)}[/tex]

[tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex]

The percentage uncertainty in his calculated value of density is:

[tex]\%e = \frac{\Delta \rho}{\rho}\times 100\,\%[/tex] (7)

If we know that [tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex] and [tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex], then the percentage uncertainty is:

[tex]\%e = \frac{\pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}} }{78.941\times 10^{-3}\,\frac{g}{mm^{3}} }\times 100\,\%[/tex]

[tex]\%e = \pm 0.713\,\%[/tex]

The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].

Answer:

A = (27.95 N, 21 N)

Explanation:

The polar co-ordinates are given as:

(r,θ) = (35 N, 37°)

Now, to convert this into polar co-ordinates (x, y), we will use following relations:

r² = x² + y²

(35)² = x² + y²

1225 = x² + y²  ----------- equation (1)

and

tan θ = y/x

tan 37° = y/x

y =  0.753 x   ------------------- equation (2)

Substituting this value in equation (1):

1225 = x² + (0.753 x)²

1225 = 1.567 x²

x² = 1225/1.567

x = √781.32

x = 27.95 N

using this value in equation (2)

y = (0.753)(27.95 N)

y = 21 N

Therefore, the vector can be represented in polar co-ordinates as:

A = (27.95 N, 21 N)

Answer:

The velocity the stone must be thrown with is 20 m/s.

Explanation:

Given;

time taken for the stone to reach it highest point, t = 2 s

acceleration due to gravity, g = 10 m/s²

The maximum height reached by the stone at the given time is calculated as;

[tex]h = \frac{1}{2}gt^2\\\\h = \frac{1}{2}*10*2^2\\\\h = 20 \ m[/tex]

Determine the initial of the stone;

v² = u² + 2gh

where;

v is the final velocity of the stone at maximum height = 0

u is the initial velocity of the stone = ?

0 = u² - 2(10 x 20)

0 = u² - 400

u² = 400

u = √400

u = 20 m/s

Therefore, the velocity the stone must be thrown with is 20 m/s.

Answer:The answer is A

Explanation: You can instantly eliminate b and d because an ion isn't mentioned, and it is impossible to have a carbon dioxide atom. C is incorrect because there has to be two oxygen's to create a carbon dioxide molecule.

Answer:

D. A carbon dioxide molecule

Explanation:

Answer:

Speed is a description of how fast an object moves; velocity is how fast and in what direction it moves. In physics, velocity is speed in a given direction. When we say a car travels at 60 km/h, we are specifying its speed.

Answer:

Workdone = 465766038 Joules.

Explanation:

Given the following data;

Mass = 1167

Initial velocity = 10m/s

Final velocity =28m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

W = ½MVf² - ½MVi²

Substituting into the equation

W = ½(1167)*28² - ½(1167)*10²

W = ½ * 1361889* 784 - ½ * 1361889 * 100

W = 533860488 - 68094450

Workdone = 465766038 Joules.