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4 years ago

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At a place where an object is thrown vertically downward with a speed of while a different object is thrown vertically upward wi

th a speed of Which object undergoes a greater change in speed in a time of 2 s?

1 answer:

Bumek [7]4 years ago

4 0

Answer:

Both objects will undergo the same change in velocity

Explanation:

m = Mass of the Earth = 5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of object

Any object which is falling has only the acceleration due to gravity.

$ma=\dfrac{GMm}{r^2}\\\Rightarrow a=\dfrac{GM}{r^2}\\\Rightarrow a=\dfrac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{(6.371\times 10^6)^2}\\\Rightarrow a=9.81364\ m/s^2$

The acceleration due to gravity on Earth is 9.81364 m/s²

So, the speeds of the objects will change at an equal rate of 9.81364 m/s² but the change will be negative when an object is thrown up.

Hence, both objects will undergo the same change in velocity.

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A 12.0-g rock is placed in a slingshot with a spring constant of 200 N/m. The rock is stretched back 0.500 m. After the slingsho

kaheart [24]

The maximum velocity of the rock is 64.55 m/s

Answer: Option B

<u>Explanation:</u>

The possessed energy by the objects while motion called kinetic energy. It is directly proportionate to mass and square of velocity. The stored energy in the system called potential energy and expressed as below,

$\text { potential energy }=\frac{1}{2} \times(\text { spring constant }) \times\left(\text { distance from equilibrium) }^{2}\right.$

$U=\frac{1}{2} \times k \times x^{2}$

U = spring’s potential energy in certain place

k = the spring constant, in N/m.

x = the spring’s distance is stretched or compressed away from equilibrium

Conservation of energy states energy neither created nor destroyed but change from one form to other. So, using this concept, equating both potential and kinetic energies equation we get,

$\frac{1}{2} \times m \times v^{2}=\frac{1}{2} \times k \times x^{2}$

Given:

m = 12 g = 0.012 kg

k = 200 N/m

x = 0.500 m

Substitute these values we get,

$\frac{1}{2} \times 0.012 \times v^{2}=\frac{1}{2} \times 200 \times(0.500)^{2}$

$v^{2}=\frac{200 \times 0.25}{0.012}=\frac{50}{0.012}=4166.66$

Taking square root, we get,

$v=64.55 \mathrm{m} / \mathrm{s}$

7 0

4 years ago

on a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.8 cm and diameter 2.05 cm fr

emmainna [20.7K]

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.8 cm and diameter 2.05 cm from a storage room to a machinist, the weight of rod w is 23.41 N .

<h3>How is the weight calculated?</h3>

Given ,

density (d) = 7800 kg/m³

D = 2.05 cm = 0.0205 m

r = D/2 = 0.0205 / 2 =0.01025 m

h = 92.8 cm = 0.928 m

The weight is calculated by ,

W = m × g

But, to find mass (m)

m = d × V

The volume of the rod, because is cylindrical, is,

V = π × r² × h

∴ V = 3.14 × (0.01025)² × 0.928

∴ V = 3.14 × 1.051 × 10⁻⁴ × 0.928

∴ V = 3.0625 × 10⁻⁴

Now to calculate mass,

m = d × V

∴ m = 7800 × 3.0625 × 10⁻⁴

∴ m = 2.3887 Kg

Now, for calculating weight

W = m × g

∴ W = 2.3887 × 9.8

∴ W = 23.41 N

∴ The weight of the rod W assuming the free-fall acceleration is 23.41 N .

<h3>What is weight?</h3>

The weight of an object is the force exerted on it by gravity.
It is define weight as a vector quantity, the force of gravity acting on an object.
Some define weight as a scalar quantity that is the magnitude of gravity.

Can learn more about weight calculation from brainly.com/question/15685221

#SPJ4

6 0

1 year ago

Speedy Sue, que conduce a 30.0 m/s, entra a un túnel de un carril. En seguida observa una camioneta lenta 155 m adelante que se

Romashka [77]

Answer:

Si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

Explanation:

Supongamos que el vehículo de Speedy Sue decelera a razón constante, mientras que la camioneta se desplaza a velocidad constante. Se requiere conocer si ambos vehículos colisionarán, lo cual implica conocer si existe algún instante tal que ambos tengan la misma posición. Consideremos además que la posición de referencia se encuentra en la posición inicial de Sppedy Sue. Entonces, las ecuaciones cinemáticas son:

Speedy Sue

$x_{S} = x_{S,o}+v_{S,o}\cdot t+\frac{1}{2}\cdot a_{S}\cdot t^{2}$

Camioneta lenta

$x_{C} = x_{C,o} +v_{C}\cdot t$

Donde:

$x_{S,o}$ , $x_{C,o}$ - Posiciones iniciales de Speedy Sue y la camioneta lenta, medidas en metros.

$v_{S,o}$ - Velocidad inicial de Speedy Sue, medida en metros por segundo.

$v_{S}, v_{C}$ - Velocidades actuales de Speedy Sue y la camioneta lenta, medidas en metros por segundo.

$a_{S}$ - Deceleración de Speedy Sue, medida en metros por segundo cuadrado.

$t$ - Tiempo, medido en segundos.

Si conocemos que $x_{S} = x_{C}$ , $x_{S,o} = 0\,m$ , $x_{C,o} = 155\,m$ , $v_{S,o} = 30\,\frac{m}{s}$ , $v_{C} =-5\,\frac{m}{s}$ y $a_{S} = -2\,\frac{m}{s^{2}}$ , encontramos la siguiente función cuadrática:

$155\,m + \left(-5\,\frac{m}{s} \right)\cdot t = 0\,m+\left(30\,\frac{m}{s} \right)\cdot t +\frac{1}{2}\cdot (-2\,\frac{m}{s^{2}} )\cdot t^{2}$

$-t^{2}+35\cdot t-155 = 0$ (Ec. 1)

Las raíces de esta función son:

$t_{1}\approx 29.798\,s$ , $t_{2} \approx 5.201\,s$

La colisión ocurriría en la raíz positiva más pequeña, es decir:

$t \approx 5.201\,s$

Ahora, la posición en que ocurriría la colisión se determina a partir de la ecuación de desplazamiento de la camioneta lenta, es decir: ( $v_{C,o} = -5\,\frac{m}{s}$ , $x_{C,o} = 155\,m$ , $t \approx 5.201\,s$ )

$x_{C} = 155\,m +\left(-5\,\frac{m}{s}\right)\cdot (5.201\,s)$

$x_{C} = 128.995\,m$

En síntesis, si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

0 0

3 years ago

An engine performs 2700 J of work on a scooter. The scooter and rider have a combined mass of 150 kg. If the scooter started at

Finger [1]

Due to the principle of conservation of energy, the work done by the engine to move the scooter converts into kinetic energy of the scooter:
$W=K= \frac{1}{2} M v^2$
where M is the combined mass of scooter and rider, and v is the velocity of the scooter. Therefore, we can find the velocity as:
$v= \sqrt{ \frac{2W}{M} } = \sqrt{ \frac{2\cdot 2700 J}{150 Kg} }=6 m/s$

6 0

4 years ago

Which of Newton's laws explains why satellites need very little fuel to stay in oribit?

Angelina_Jolie [31]

Sattelites don't need any fuel to stay in orbit. The applicable law is...."objects in motion tend to stay in motion". Having reached orbital velocity, any such object is essentially "falling" around the earth. Since there is no (or at least very little) friction in the vacuum of space, the object does not slow.... It simply continues.

Sattelites in "low" earth orbit do encounter some friction from the very thin upper atmosphere, and they will eventually "decay".

:)

4 0

3 years ago