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2 years ago

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1. A rocket is launched from a 300 cm rail. The upper Launch Lug is placed 1 point 150 cm from the bottom of the rocket. What is

the effective distance of the Launch Rail? 300 cm 150 cm 3 m 300 m O 1m O 1.5 m

1 answer:

patriot [66]2 years ago

4 0

Answer:

i think its going to be 150 because its half of 300

Explanation:

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a car travelling at 50km/h from rest covers a distance of 10km in 40minutes. Calculate the acceleration

Margarita [4]

Answer:

$9.67\cdot 10^{-3}m/s^2$

Explanation:

We can solve the problem by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the car in this problem:

u = 0 (it starts from rest)

$v=50 km/h \cdot \frac{1000}{3600}=13.9 m/s$ is the final velocity

s = 10 km = 10 000 m is the displacement

Solving for a, we find:

$a=\frac{v^2-u^2}{2s}=\frac{13.9^2}{2(10000)}=9.67\cdot 10^{-3}m/s^2$

7 0

3 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo

Helga [31]

Answer: $3.23 m/s^2$

Explanation:

Given

$\mu_s =0.330$

Frictional Force is balanced by force due to car acceleration

Frictional force $F_s$

$F_s=ma_{min}$

$\mu_sN=ma_{min}$

$\mu_s\cdot mg=ma_{min}$

$a_{min}=\mu_s \cdot g=0.330\times 9.8=3.23 m/s^2$

6 0

3 years ago

The density of gold is 19.3. G/cm(3). If a nugget of iron pyrite and nugget of gold each have a mass of 50 g, what can you concl

tangare [24]

The pyrite will be bigger, because its density is much lower.

I <em>do</em> know that the gold's volume will be 2.5906 (With a bunch more numbers after it)

50 divided by 19.3 = 2.5906

5 0

3 years ago

If a blender is plugged into a 168 V outlet that supplies 80.8 A of current, what amount of

VladimirAG [237]

Answer:

Power = 13.5744 kilowatts

Explanation:

Power is the rate at which work is done. ... Great power means a large amount of work or energy developed in a short time. For example, when a powerful car accelerates rapidly, it does a large amount of work and consumes a large amount of fuel in a short time.

Formula for power = work/time

= IVT/T

= IV

Where I is the current

And V is the voltage

The voltage V supply = 168 v

The current A supply = 80.8 A

Power = 80.8*168

Power = 13574.4 watts

Power = 13.5744 kilowatts

5 0

3 years ago

A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is

Soloha48 [4]

Answer: $V=5839.051m/s$

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;:

$T=1.26(10)^{4}s$ is the period of the satellite

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=5.98(10)^{24}kg$ is the mass of the Earth

$a$ is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity $V$ is given by:

$V=\sqrt{\frac{GM}{a}}$ (2)

Now, from (1) we can find $a$ , in order to substitute this value in (2):

$a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}$ (3)

$a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}$ (4)

$a=11705845.57m$ (5)

Substituting (5) in (2):

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}$ (6)

$V=5839.051m/s$ (7) This is the speed at which the satellite travels

6 0

3 years ago