Which of the following is the most likely global climate change?

Z. A force that gives a 8-kg objet an acceleration of 1.6 m/s^2 would give a 2-kg object an

Paladinen [302]

Answer:

$\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}$

Explanation:

We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.

<h3>1. Force of 1st Object </h3>

First, we can find the force of the first, 8 kilogram object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F=m \times a$

The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.

m= 8 kg
a= 1.6 m/s²

Substitute these values into the formula.

$F= 8 \ kg * 1.6 \ m/s^2$

Multiply.

$F= 12.8 \ kg*m/s^2$

<h3>2. Acceleration of the 2nd Object </h3>

Now, use the force we just calculated to complete the second part of the problem. We use the same formula:

$F= m \times a$

This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.

F= 12.8 kg *m/s²
m= 2 kg

Substitute the values into the formula.

$12.8 \ kg*m/s^2= 2 \ kg *a$

Since we are solving for the acceleration, we must isolate the variable (a). It is being multiplied by 2 kg. The inverse of multiplication is division. Divide both sides of the equation by 2 kg.

$\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}$

$\frac {12.8 \ kg*m/s^2}{2 \ kg}=a$

The units of kilograms cancel.

$\frac {12.8}{2}\ m/s^2=a$

$6.4 \ m/s^2=a$

The acceleration is 6.4 meters per square second.

4 0

2 years ago

P6: An object of mass m sits on a spring of constant k in an elevator that is accelerating upwards with acceleration a. a) In te

tankabanditka [31]

Answer:

(a). The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). The acceleration is 1.5 g.

Explanation:

Given that,

Acceleration = a

mass = m

spring constant = k

(a). We need to calculate the spring compressed

Using balance equation

$kx-mg=ma$

$x=\dfrac{ma+mg}{k}$ ....(I)

The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,

The compression is given by

$x=2.5\times x_{0}$

Here, acceleration is zero

So, $x=2.5\times\dfrac{mg}{k}$

We need to calculate the acceleration

Put the value of x in equation (I)

$2.5\times \dfrac{mg}{k}=\dfrac{ma+mg}{k}$

$2.5\times\dfrac{mg}{k}=\dfrac{m}{k}(a+g)$

$a=2.5g-g$

$a=1.5g$

Hence, (a). The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). The acceleration is 1.5 g.

8 0

3 years ago

How many joules of heat would be required to heat 0.5 kg of aluminum by 2 kelvin

melisa1 [442]

0.902 joules of energy

4 0

3 years ago

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ν0. find the minimum en

shusha [124]

The energy carried by the incident light is
$E=hf$
where h is the Planck constant and f is the frequency of the light. The threshold frequency is the frequency that corresponds to the minimum energy needed to eject the electrons from the metal, so if we substitute the threshold frequency in the formula, we get the minimum energy the light must have to eject the electrons:
$E=hf=(6.63 \cdot 10^{-34}Js)(5.64 \cdot 10^{14}s^{-1})=3.74 \cdot 10^{-19}J$

4 0

4 years ago

On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m

Nastasia [14]

Answer:

$F = 1.489*10^{-7} N$

Explanation: Weight of space probes on earth is given by: $W= m*g$

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 $\frac{m}{s^{2} }$ )

Therefore,

$m_{1} = \frac{14500}{9.81}$

$m_{1} = 1478.08 kg$

Similarly,

$m_{2} = \frac{4800}{9.81}$

$m_{2} = 489.29 kg$

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

$F = \frac{Gm_{1} m_{2}}{R^{2} }$

G= gravitational constant ( $6.67 * 10^{-11} m^{3} kg^{-1} s^{-2}$ )

$m_{1} , m_{2}$ = masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

$F = 1.489*10^{-7} N$

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

7 0

3 years ago