Please please answer click on image to see full pic

The pressure in an automobile tire is 198 kpa at 300.0 k. at the end of a trip on a hot, sunny day, the pressure has risen to 22

Kamila [148]

We can use the ideal gas equation which is expressed as PV = nRT. At a constant volume and number of moles of the gas the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:

T1/P1 = T2/P2

T2 = P2 x T1 / P1

T2 = 225 x 300 / 198

3 0

4 years ago

Will mark the brainliest

Gennadij [26K]

Answer:

OK draw a diagram - you have the force from (b) acting sown the slope and a component of the weight. Use F=ma to get the deceleration and then use SUVAT. Post your working if this doesn't work.

friction = 900N

braking = 3100N

total = 4000N

w= mxg = 1000*9.81= 9810N

total = 9810 + 4000 = 13810N

force/mass = 13810/1000 = 13.81ms^-2

then using v^2 = u^2 + 2as, i get s as 14 but it is incorrect

Explanation:

4 0

3 years ago

(a) What is the escape speed on a spherical asteroid whose radius is 500. km and whose gravitational acceleration at the surface

navik [9.2K]

Answer:

a) v= 1732.05m/s

b) d=250000m

c) v= 1414.214m/s

Explanation:

Notation

M= mass of the asteroid

m= mass of the particle moving upward

R= radius

v= escape speed

G= Universal constant

h= distance above the the surface

Part a

For this part we can use the principle of conservation of energy. for the begin the initial potential energy for the asteroid would be $U_i =-\frac{GMm}{R}$ .

The initial kinetic energy would be $\frac{1}{2}mv^2$ . The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

$-\frac{GMm}{R}+\frac{1}{2}mv^2=0$ (1)

Dividing both sides by m and replacing $\frac{GM}{R}$ by $a_g R$

And the equation (1) becomes:

$-a_g R+\frac{1}{2} v^2=0$ (2)

If we solve for v we got this:

$v=\sqrt{2 a_g R}=\sqrt{2x3\frac{m}{s^2}x500000m}=1732.05m/s$

Part b

When we consider a particule at this surface at the starting point we have that:

$U_i=-\frac{GMm}{R}$

$K_i=\frac{1}{2}mv^2$

Considering that the particle is at a distance h above the surface and then stops we have that:

$U_f=-\frac{GMm}{R+h}$

$K_f=0$

And the balance of energy would be:

$-\frac{GMm}{R}+\frac{1}{2}mv^2 =-\frac{GMm}{R+h}$

Dividing again both sides by m and replacing $\frac{GM}{R}$ by $a_g R^2$ we got:

$-a_g R+\frac{1}{2}v^2 =-\frac{a_g R^2}{R+h}$

If we solve for h we can follow the following steps:

$R+h=-\frac{a_g R^2}{-a_g R+\frac{1}{2}v^2}$

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

$h=\frac{2a_g R^2}{2a_g R-v^2}-R$

Replacing:

$h=\frac{2x3\frac{m}{s^2}(500000m)^2}{2(3\frac{m}{s^2})(500000m)-(1000m/s)^2}- 500000m=250000m$

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

$U_i=-\frac{GMm}{R+h}$

$K_i=0$

And after the particle reach the asteroid we have this:

$U_f=-\frac{GMm}{R}$

$K_f=\frac{1}{2}mv^2$

So the balance of energy would be:

$-\frac{GMm}{R+h}=-\frac{GMm}{R}+\frac{1}{2}mv^2$

Replacing again $a_g R^2$ instead of GM and dividing both sides by m we have:

$-\frac{a_g R^2}{R+h}=-a_g R+\frac{1}{2}v^2$

And solving for v:

$a_g R-\frac{a_g R^2}{R+h}=\frac{1}{2}v^2$

Multiplying both sides by two and taking square root:

$v=\sqrt{2a_g R-\frac{2a_g R^2}{R+h}}$

Replacing

$v=\sqrt{2(3\frac{m}{s^2})(500000m)-\frac{2(3\frac{m}{s^2}(500000m)^2}{500000+1000000m}}=1414.214m/s$

3 0

3 years ago

A rocket is launched straight up from Earth's surface at a speed of 15,000 m/s. What is its speed when it is very far away from

Colt1911 [192]

Answer:

1/2 m v2^2 = 1/2 m v1^2 - G M m / R conservation of energy

v1^2 - v2^2 = 2 G M / R rearranging terms

2 G M / R = 2 * 6.67 * 5.98 / 6.37 E7 = 1.25E8

v1^2 - v2^2 = 1.25E8

v2^2 = v1^2 - 1.25E8 = (1.5^2 - 1.25) * E8

v2 = 1.00E4 = 10,000 m/s

3 0

2 years ago

A pair of closely spaced slits is illuminated with 650.0-nm light in a Young's double-slit experiment. During the experiment, on

Masteriza [31]

Answer:

The minimum thickness of the Lucite plate is 0.670 μm.

Explanation:

Given that,

Wavelength = 650.0 nm

Index of refraction = 1.485

We need to calculate the minimum thickness of the Lucite plate

Using formula of thickness

$t(n-1)=\dfrac{\lambda}{2}$

$t=\dfrac{\lambda}{2(n-1)}$

Where, n = Index of refraction

$\lambda$ = wavelength

Put the value into the formula

$t=\dfrac{650.0\times10^{-9}}{2(1.485-1)}$

$t =0.670\ \mu m$

Hence, The minimum thickness of the Lucite plate is 0.670 μm.

7 0

3 years ago

Read 2 more answers