A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds.
Determine the acceleration of the car and the distance traveled.

Answer:

0.364

I believe... Good luck!

Answer:

45 degrees

Explanation:

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

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:

1144.8

Explanation:

1 meter per second is equal to 3.6 kilometers per hour

Answer:

Explanation:

The area of the base of the statue can be found by using the formula

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

f is the force

p is the pressure

From the question we have

[tex]a = \frac{1000}{20000} = \frac{1}{20} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

C

Explanation:

Answer:

See below

Explanation:

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( hope it helped <3 )

Answer:

Explanation:

Given

the range = 5.00m (distance moved in horizontal direction)

Height = 0.150m

Required

Initial velocity of the dart

Using the formula for calculating range of a projectile;

R = U√2H/g

5 = U√2(0.15)/9.8

5 = U√0.0306

5 = 0.1749U

U = 5/0.1749

U = 28.59m/s

Hence the initial horizontal velocity of the dart is 28.59m/s

Answer:

28.59

Explanation:

khan academy

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

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.

Answer:55,000 kg•m/s

Explanation:

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:

it is hit by an external force

Explanation:

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:

ask questions that the findings bring up

Explanation:

hope this helps :)

Answer:

ask questions that the findings bring up

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:

Replication of the DNA must occur. Segregation of the "original" and its "replica" follow. Cytokinesis ends the cell division process. Whether the cell was eukaryotic or prokaryotic, these basic events must occur.

Answer:

I think it's A and D.

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.

Learn more about projectile.

brainly.com/question/11422992

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Answer:

no

Explanation:

because it will bind together in the molecular level

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:

B. An opposing force caused by friction produced a lower acceleration than calculated

Explanation:

Answer:

An opposing force caused by friction produced a lower acceleration than calculated.

Explanation:

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.

Learn more here:https://brainly.com/question/21866017

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.

Answer:

C. It will remain constant because the total amount of thermal energy  plus chemical potential energy stays the same.

Explanation:

According to the law of conservation of energy, the total energy of this system will remain constant because the total amount of thermal energy plus chemical potential energy stays the same.

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].