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RideAnS [48]
3 years ago
5

please answer 1-3 and use a separate piece of paper thank you i've been trying those problems for 3 day.... ikr Sad

Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
4 0
\frac{total distance}{ total time}

\frac{72m}{36s} = 2m/s

Same thing for the rest 
 2) 5km/h
 3)360km/h

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Is being social media famous worth real-life relationships?
Maru [420]

Answer:

No

Explanation:

Social media fame isnt worth it, because youre not really able to connect with people. Real life relationships with friends, family, and everything else is way more important. having people notice you or be attracted to you through the internet is nothing compared to real life social interactions. Dont leave people just to be online, youre going to regret it.

3 0
3 years ago
Mark each of the following statementsas true or false. If a statement refers to "two objects"interacting via some force, you are
slega [8]

Answer:

Explanation:

A,D,E

8 0
3 years ago
The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu
BartSMP [9]

Answer:

hello your question is incomplete  attached below is the missing part  

answer : short period oscillations frequency  = 0.063 rad / sec

              phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= ( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns}  ) = 0

where :

w_{np}  = Natural frequency of plugiod oscillation

\alpha _{p} = damping ratio of plugiod  oscilations

comparing the general form with the given equation

w^{2} _{np}  = 18.2329

w^{2} _{ns} = 0.003969

hence the short period oscillation frequency ( w_{ns} ) =  0.063 rad/sec

phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

8 0
3 years ago
A scientist is studying a sample of matter the matter has no definite shape or volume and it can conduct an electric current. Th
Anna35 [415]
D. plasma ..................
4 0
3 years ago
Read 2 more answers
A solid conducting sphere of radius 2.00 cm has a charge of 8.84 μC. A conducting spherical shell of inner radius 4.00 cm and ou
AlladinOne [14]

Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5

Explanation:  In order to solve this problem  we have to use the gaussian law in the mentioned regions.

Region 1; 0<r<2

∫E.ds=Qinside the gaussian surface/ε0

inside of the solid conducting sphere the elevctric field is zero because the charge is located at the surface on this sphere.

Region 2; 2<r<4;

E.4*π*r^2=8,84/ε0

E=8,84/(4*π*ε0*r^2)

Region 3; 4<r<5

E=0 because is inside the conductor.

Finally

Region 4; r>5

E.4*π*r^2=(8,84-2.02)/ε0

6 0
3 years ago
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