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Monica [59]
2 years ago
10

Help plss i just need it plss​

Physics
1 answer:
Travka [436]2 years ago
4 0

Answer:

ooh thanks but don't give me advise

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PLS HELP ME. A 0.0780 kg lemming runs off a 5.36m high cliff at 4.84 m/s what is it potential energy when it lands?​
Pepsi [2]

Answer:

p.e=0.078kg×1/2×5.36m

p.e=0.913j

4 0
3 years ago
An airplane flies in a loop (a circular path in a vertical plane) of radius 160 m . The pilot's head always points toward the ce
notka56 [123]

Answer:

a) 39.6 m/s b) 4123 N

Explanation:

a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).

Fnet=ma

ma=m(v^2/R) (centripetal acceleration)

mg=m(v^2/R)

m cancels out (this is why pilot feels weightless) so,

g=(v^2/R)

9.8 m/s^2 = v^2/160 m

v^2=1568 m^2/s^2

v=39.6 m/s

b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.

Convert 300 km/hr to m/s

300 km/hr=83.3 m/s

Convert pilot's weight into mass:

760 N = 77.55 kg

Fnet=ma

n-mg=m(v^2/R)

n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)

n=3363.2 N+760 N=4123 N

5 0
2 years ago
Each orbit is limited to a maximum of four electrons true or false
Naya [18.7K]
This question would be false
7 0
3 years ago
Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

Here

r=e^u\\r=e^{\pi/4}\\r=2.1932 m

r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m

So

a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

So

a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

The radial component of acceleration is 8.77 m/s^2

6 0
3 years ago
What is the volume of a rock with a density of 3.00 g/cm3 and a mass of 600g?
Mademuasel [1]
The equation of D = m/V

Where D = density
m = mass
and V = volume

We are solving for V, so with the manipulation of variables we multiply V on both sides giving us 
V(D) = m 
now we divide D on both sides giving us
V = m/D 

We know our mass which is 600g and our density is 3.00 g/cm^3
so
V = 600g/3.00g/cm^3 = 200cm^3  or 200mL

a cubic centimeter (cm^3) is one of the units for volume. It's exactly like mL. 1 cm^3 = 1 mL
 
If you wish to change it to L, you'd have to convert. 
5 0
2 years ago
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