Which scenario is an example of a physical change?

The palm of sheilas hand has an area of 0.0017 m2. If the atmospheric pressure on the palm is 100,000 Pa, what force is being ex

Serjik [45]

Answer:

170 N

Explanation:

F=pressure(pa) ÷ area(m^2)

=100.000 ÷ 0,0017

=170 N

8 0

2 years ago

Sobre un barco, que se mueve en forma rectilínea, y con velocidad constante de 30 [km/h], se mueve un perro en el mismo sentido

almond37 [142]

Answer:

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

Explanation:

Para determinar la velocidad del perro con respecto al observador sentado desde la playa a través del concepto de velocidad relativa, descrito en la siguiente fórmula:

$v_{P/B} = v_{P} - v_{B}$ (1)

Donde:

$v_{P/B}$ - Velocidad del perro relativo al barco, en kilómetros por hora.

$v_{P}$ - Velocidad del perro con respecto al observador, en kilómetros por hora.

$v_{B}$ - Velocidad del barco con respecto al observador, en kilómetros por hora.

Si sabemos que $v_{B} = 30\,\frac{km}{h}$ y $v_{P/B} = 10\,\frac{km}{h}$ , entonces la velocidad del perro con respecto al observador es:

$v_{P} = v_{B} + v_{P/B}$

$v_{P} = 30\,\frac{km}{h} + 10\,\frac{km}{h}$

$v_{P} = 40\,\frac{km}{h}$

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

6 0

3 years ago

if a body of mass 2kg moving with Velocity of 15m/s collides with a stationary body of same mass then after elastic collision 2n

pochemuha

Answer:

v = 15 [m/s]

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is equal to the product of mass by Velocity.

$P=m*v$

where:

P = linear momentum [kg*m/s]

m = mass = 2 [kg]

v = velocity = 15 [m/s]

$P=2*15\\P=30 [kg*m/s]$

Since we know that momentum is conserved, that is, all momentum is transferred to the second body, we can determine the velocity of the second body, since the mass is equal to that of the first body.

$30=2*v\\v = 30/2\\v = 15 [m/s]$

5 0

3 years ago

An automobile with a mass of 1180 kg is traveling at a speed v = [06] ____________________ m/s. a. What is its kinetic energy in

iogann1982 [59]

Complete Question:

An automobile with a mass of 1180 kg is traveling at a speed v =2.51 m/s. What is its kinetic energy in SI units? What speed (m/s) must an 82.7-kg person move to have the same kinetic energy? At what speed (m/s) must is 12.1-g bullet move to have the same kinetic energy? What would be the speed (m/s) of the automobile if its kinetic energy were doubled?

Answer:

a) 3717.1 J b) 9.48 m/s c) 783.8 m/s d) 3.55 m/s

Explanation:

a)

By definition, the kinetic energy of a mass m with a speed v, is as follows:

$K = \frac{1}{2} * m *v^{2}$

if m= 1180 Kg, and v= 2.51 m/s, the kinetic energy can be calculated as follows:

$K = \frac{1}{2} * m *v^{2} = \frac{1}{2} * 1180 kg*(2.51 m/s)^{2} = 3717.1 J$

b)

If the kinetic energy must be the same, and m= 82,7 Kg, we can write the following expression:

$K = \frac{1}{2} * m *v^{2} = \frac{1}{2} * 82.7 kg*((v)(m/s))^{2} = 3717.1 J$

We can solve the above equation as follows:

$v =\sqrt{\frac{2*K}{m} } = \sqrt{\frac{2*3717.1J}{82.7kg} } = 9.48 m/s$

c)

If K remains the same, and m = 12.1 g = 0.0121 kg (in SI units). we can solve for v as follows:

$v =\sqrt{\frac{2*K}{m} } = \sqrt{\frac{2*3717.1J}{0.0121kg} } = 783.8 m/s$

d)

Now, if the kinetic energy were doubled, we would have the following equation:

$K = \frac{1}{2} * m *v^{2} = \frac{1}{2} * 1180 kg*((v) m/s)^{2} = 3717.1 J * 2 = 7434.2 J$

We can solve for the new speed v as follows:

$v =\sqrt{\frac{2*K}{m} } = \sqrt{\frac{2*7434.2J}{1180kg} } = 3.55 m/s$

8 0

3 years ago

A faulty thermometer reads 2°C when dipped in ice at 0°C and 95°C when dipped in steam at 100°C. What would this thermometer rea

kvasek [131]

Answer:

The read will be 20.9[C] $x=\frac{(y-y_{1} )}{(y_{2} -y_{1} )}*(x_{2}-x_{1})+y_{1}\\ x=\frac{(18-0 )}{(95 -0 )}*(100-2)+2\\\\x= 20.9[C]$

Explanation:

This is a problem related to linear interpolation, Linear interpolation consists of tracing a line through two known points y = r (x) and calculating the intermediate values according to this line.

The equation of a known line two points (x1, y1)and (x2, y2) = (2,0) (100,95) is:

$\frac{(y-y_{1} )}{(y_{2} -y_{1} )}=\frac{(x-y_{1} )}{(x_{2}-x_{1} )}$

If we clear y from the equation we have:

[tex]y=\frac{(x-y_{1})}{(x_{2}-x_{1} )}*(y_{2} -y_{1} )+y_{1} \\replacing

4 0

3 years ago