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3 years ago

Helpppp please I'm really stuck on this question

Physics

2 answers:

finlep [7]3 years ago

8 0

Your answer would be C. We just went over this in my class

amm18123 years ago

6 0

Is this a test? If it is, i’m not allowed to help as i would be seen as a bad person! I hope you pass, god bless.

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A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.

Genrish500 [490]

<u>Answer:</u>

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

<u>Explanation:</u>

a) The height of ramp = 1.5 meter

Horizontal distance he must clear = 22 meter

The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

We have equation of motion, $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when displacement = 1.5 meter.

$1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds$

So the car has to cover a distance of 22 meter in 2.119 seconds.

So minimum speed required = 22/0.553 = 39.78 m/s

Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 $m/s^2$

Substituting

$-1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0$

In case of horizontal motion

Horizontal speed of car = V cos 7 = 0.993V

So it has to travel 22 meter in t seconds

0.993Vt = 22, Vt = 22.155 m

Substituting in the equation $4.9t^2-0.122Vt-1.5=0$

We will get $4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds$

Speed required = 22.155/0.926 = 23.93 m/s

Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

7 0

4 years ago

Three bulbs are connected by tubing, and the tubing is evacuated. The volume of the tubing is 45.0 mL. The first bulb has a volu

kipiarov [429]

Answer:

The final pressure of the whole system is 34.80 atm.

Explanation:

Given that,

Volume = 45.0 ml

Volume of first bulb = 77.0 mL

Pressure = 8.89 atm

Volume of second bulb = 250 mL

Pressure = 2.82 atm

Volume of third bulb = 21.0 mL

Pressure = 8.42 atm

We need to calculate the final pressure of the whole system

Using formula of pressure

$P_{1}V_{1}+P_{2}V_{2}+P_{3}V_{3}+P_{t}V_{t}=P_{f}V_{f}$

Where, $P_{1}$ = pressure of first bulb

$P_{2}$ = pressure of second bulb

$P_{3}$ = pressure of third bulb

$P_{4}$ = initial pressure of tube

$V_{1}$ = Volume of first bulb

$V_{2}$ =Volume of second bulb

$V_{3}$ = Volume of third bulb

$V_{4}$ = Initial volume of tube

Put the value into the formula

$8.89\times77.0+250\times2.82+21.0\times8.42+0=P_{f}\times45$

$P_{f}=\dfrac{1566.35}{45}$

$P_{f}=34.80\ atm$

Hence, The final pressure of the whole system is 34.80 atm.

4 0

3 years ago

The atomic mass of an atom equals the sum of he masses of the ___.

Aleksandr-060686 [28]

The atomic mass of an atom ... in Atomic Mass Units ... is the sum of the numbers of protons and neutrons in its nucleus.

3 0

3 years ago

The chemical equation below shows the reaction of methane (CH4) and oxygen gas (O2) to form carbon dioxide (CO2) and water (H2O)

Nuetrik [128]

<h2>Answer </h2>

Option C - Carbon Dioxide and Water

<u>Explanation </u>

The chemical equation reaction of methane (CH4) and oxygen gas (O2) to form carbon dioxide (CO2) and water (H2O).

CH4 + 2O2 → CO2 + 2H2O.

In this equation, carbon dioxide and water are products. Products are substances which are formed after a chemical reaction so in this case carbon dioxide and water are formed as a result after a chemical reaction. Therefore, these are products of reaction of methane and oxygen gas.

8 0

3 years ago

Read 2 more answers

how much energy is needed to raise the temperature of 2kg of copper from 0°c to 10°c. the specific heat capacity of copper is 38

kogti [31]

Explanation:

H= mass× specific heat capacity×change in temperature

=2×380×(10-0)

=2×380×10

=7600Joules

4 0

3 years ago