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2 years ago

A dog runs with an initial velocity of 7.5 m/s on a waxed floor. It slides to a stop in 15 seconds. What is the acceleration?

Physics

1 answer:

Leni [432]2 years ago

3 0

Answer:

-0.5 m/s^2

Explanation:

Acceleration = change in velocity / total time

7.5 / 15 = 0.5

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A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____

Rama09 [41]

Answer:

Magnification, m = -0.42

Explanation:

It is given that,

Height of diamond ring, h = 1.5 cm

Object distance, u = -20 cm

Radius of curvature of concave mirror, R = 30 cm

Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)

Using mirror's formula :

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$ , f = focal length of the mirror

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{-20}$

v = -8.57 cm

The magnification of a mirror is given by,

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-8.57)}{-20}$

m = -0.42

So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.

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3 years ago

A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be

stepan [7]

Answer:

$p_2 = 664081 N/m^{2}$

Explanation:

from the ideal gas law we have

PV = mRT

$P = \rho RT$

$\rho = \frac{P}{RT}$

HERE R is gas constant for dry air = 287 J K^{-1} kg^{-1}

$\rho = \frac{7.00 10^{5}}{287(18+273)}$

$\rho = 8.38 kg/m^{3}$

We know by ideal gas law

$\rho = \frac{m_1}{V_1}$

$m_1 = \rho V_1 = 8.38 *2*10^{-3}$

$m_1 = 0.0167 kg$

for m_2

$m_2 = \rho V_i - V_removed$

$m_2 = 8.38*(.002 - 10^{-4})$

$m_2 = 0.0159 kg$

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

$\frac{p_2}{p_1}=\frac{m_2}{m_1}$

$p_2 = p_1 * \frac{m_2}{m_1}$

$p_2 =7*10^{5} * \frac {.0159}{0.0167}$

$p_2 = 664081 N/m^{2}$

8 0

2 years ago

Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay

kaheart [24]

Answer:

0.278 m/s

Explanation:

We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.

So we can write:

$mu=(m+M)v$

where

m = 0.200 kg is the mass of the koala bear

u = 0.750 m/s is the initial velocity of the koala bear

M = 0.350 kg is the mass of the other clay model

v is their final combined velocity

Solving the equation for v, we get

$v=\frac{mu}{m+M}=\frac{(0.200)(0.750)}{0.200+0.350}=0.278 m/s$

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3 years ago