Ask question

Ask question

All categories

3 years ago

11

The four tires of an automobile are inflated to a gauge pressure of 2.2 ✕ 105 Pa. Each tire has an area of 0.022 m2 in contact w

ith the ground. Determine the weight of the automobile.

1 answer:

Trava [24]3 years ago

7 0

Answer:

$P =\frac{F}{A}$

And solving for F we got:

$F = PA$

And replacing we got:

$F = 2.2x10^5 Pa * 0.088 m^2 =19360 N$

And if we want the mass we can solve from this formula:

$F =mg$

With g=9.8 m/s^2 the gravity

Explanation:

For this case we can find the total area of the 4 tires and we got:

$A = 4*0.022 m^2 = 0.088m^2$

Now we know that the total pressure is:

$P=2.2 x10^5 Pa$

And then we can find the weight with from this formula:

$P =\frac{F}{A}$

And solving for F we got:

$F = PA$

And replacing we got:

$F = 2.2x10^5 Pa * 0.088 m^2 =19360 N$

And if we want the mass we can solve from this formula:

$F =mg$

With g=9.8 m/s^2 the gravity

You might be interested in

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

8 0

4 years ago

How far will a free falling object fall in 8.7 secs if it started from rest? Remember acceleration is negative for free fall. Do

sleet_krkn [62]

Answer:

h~=371.26m

Explanation:

when an object falls we use the equations of accelerated motion. There is only one that gives distance.

$x = ut + \frac{1}{2} a {t}^{2}$

Since we have no initial velocity (started from rest) we can get rid of the (ut) term

where a we substitute g (gravitational acceleration, constant for given heights and almost 9.81m/s^2).

$h = \frac{1}{2} g {t}^{2} = \frac{1}{2} \times 9.81 \times {8.7}^{2} = 371.26m$

4 0

3 years ago

the potential energy of a 40kg cannon ball is 14000j. How high was the cannon ball to have this much potential energy?

noname [10]

We will use the formula p = mgh p is potential energy. m is mass of object in kg g is acceleration due to gravity (9.8m/s²) h is height of the objects displacement in meters. p = mgh → mgh = p → h = p / mg p is 14000j, m is 40kg and g is 9.8 m/s² h = 14000 / 40 × 9.8 → h = 1400 / 392 → h = 35.7 Therefore , the cannonball was 35.7 meters high .

7 0

3 years ago

A woman (m = 45kg) is kneeling on the shoulders of a man (m = 70kg) in pair figure-skating while both are traveling at 1.5 m/s.

Temka [501]

Both the man and the woman will be experiencing the same impulse, due to Newton's third law of motion.

Explanation:

This problem considers law of conservation of momentum as well as Newton's third law of motion. As per the third law of motion, every action has an equal and opposite reaction. And the law of conservation of momentum states that the momentum is conserved after collision.

Since, here the woman is kneeling on the man, it is similar to inelastic collision where the man and woman are moving with same velocity. And as the man accelerates the woman, the woman will exhibit a forward force. While at the same time, the man will be experiencing a backward force of same magnitude as of the woman. This is in consistence with the third law of motion.

So both the man and woman will experience force of same magnitude but opposite in direction. As impulse is directly proportional to the force acting on any object in a given duration, so both the man and the woman will be experiencing the same impulse.

3 0

4 years ago

Answer pls will give 20 points per person

LuckyWell [14K]

Answer: question 1 is b I believe

Explanation:

every action has an opposite reaction

7 0

3 years ago

Read 2 more answers