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ArbitrLikvidat [17]
2 years ago
6

A flask with a tap has a volume of 200cm3 when full of air, tye flask has a mass of 30.98g. the flask is connected to vacuum pum

p, the air is pimped put and then the tap is closed, the flask now has mass of 30.72g
calculate:
a. the mass of air in the flask before connecting to vacuum pump, in g
b. the density of the air in the flask

​
Physics
1 answer:
fgiga [73]2 years ago
7 0
1. 30.98g
2 D = M / V = 30.98/200 = 0.1549
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Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

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3 years ago
Write two examples of machine which changes the direction of force​
Ostrovityanka [42]

Answer:

Washign machine and swingset

Explanation:

4 0
2 years ago
In 2 1/2 hours an airplane travels 1150 km against the wind. It takes 50 min to travel 450 km with the wind. Find the speed of t
Tamiku [17]

Answer:

Velocity of airplane is 500 km/h

Velocity of wind is 40 km/h

Explanation:

V_a= Velocity of airplane in still air

V_w= Velocity of wind

Time taken by plane to travel 1150 km against the wind is 2.5 hours

V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)

Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours

V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)

Subtracting the two equations we get

V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h

Applying the value of velocity of wind to the first equation

V_a-40=460\\\Rightarrow V_a =500\ km/h

∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h

5 0
3 years ago
Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
mylen [45]

Answer: 3.7 \times 10^{-4} N

Explanation:

The gravitational pull between two object is given by:

F = G\frac{Mm}{r^2}

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

4 0
3 years ago
Read 2 more answers
A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the
Liula [17]

Answer:

The average force on ball by the golf club is 340 N.

Explanation:

Given that,

Mass of the golf ball, m = 0.03 kg

Initial speed of the ball, u = 0

Final speed of the ball, v = 34 m/s

Time of contact, \Delta t=0.003\ s

We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

F=\dfrac{\Delta p}{\Delta t}\\\\F=\dfrac{mv-mu}{\Delta t}\\\\F=\dfrac{mv}{\Delta t}\\\\F=\dfrac{0.03\ kg\times 34\ m/s}{0.003\ s}\\\\F=340\ N

So, the average force on ball by the golf club is 340 N.

4 0
3 years ago
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