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3 years ago

By using_______you can determine your exact position and altitude on earth?

Physics

1 answer:

kupik [55]3 years ago

4 0

By using a ''GPS'' you can determine your exact position and altitude on earth.

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A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

$y=\sqrt{1+x^3}$

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Now at (2,3)

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s$

7 0

3 years ago

You use 42 J of energy to move a 6.0 N object in 2 seconds. How far did you move it?

bogdanovich [222]

22 that's how much I moved

7 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

Time taken to reach 1180 m is 11.29 seconds

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

Time the rocket will keep going up after the engines shut off is 13.06 seconds.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$

Time taken by the rocket to fall from maximum height is 20.29 seconds

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0

3 years ago

Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter th

denis-greek [22]

Answer:

In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force

Explanation:

7 0

3 years ago

The kinetic energy K of a moving object varies jointly with its mass m and the square of its velocity v. If an object weighing 4

nata0808 [166]

Answer:

K' = 1777.777 J

Explanation:

Given that

m = 40 kg

v= 15 m/s

K=1000

Given that kinetic energy(K) varies with mass(m) and velocity(v)

K= C(mv²)

Where

C= Constant

m=mass

v=velocity

When

m = 40 kg ,v= 15 m/s ,K=1000

K= C(mv²)

1000 = C( 40 x 15²)

C=0.111111

When m = 40 kg and v= 20 m/s

K' = C(mv²)

K= 0.1111 x (40 x 20²)

K' = 1777.777 J

5 0

3 years ago

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