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PtichkaEL [24]
3 years ago
10

A bird accelerating from rest at a constant rate,experiences a displacement of 28 m in 11s.What is its acceleration

Physics
1 answer:
Ksenya-84 [330]3 years ago
4 0
Use the formula below for this question:

v_{f} = v_{i} + 2 * a * d

re-arrange to solve for a:

a = \frac{v_{f} - v_{i}}{2 * d}

now simply plug in your variables and there's your answer :). If you ever get stuck, you can look up the kinematic equations!
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The Sears Tower vibrates back and forth, it makes about 8.6
Leni [432]

Answer: Frequency is 0.143 Hz; Period is 7 seconds

Explanation:

Number of vibrations = 8.6

Time required = 60 seconds

Period (T) = ?

Frequency of the vibrations (F) = ?

A) Recall that frequency is the number of vibrations that the Sears tower completes in one second.

i.e Frequency = (Number of vibrations / time taken)

F = 8.6/60 = 0.143Hz

B) Period, T is inversely proportional to frequency. i.e Period = 1/Frequency

T = 1/0.143Hz

T = 7 seconds

Thus, the frequency and period of the vibrations of the Sears Tower are 0.143 Hz

and 7 seconds respectively.

7 0
4 years ago
When driving at night, only use your high-beam headlights___
zzz [600]

Answer:

B

Explanation:

While answer C may sound correct, Answer B is makes more sense. We know you cant use High-beam lights when u cant see ongoing traffic because it could affect the other driver coming across from you. Its good to use it when legal and safe, but in that term I still don't believe there's no reason for HIGH-beamed. That's this leaves B, when you are on u lighted streets.

6 0
3 years ago
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
4 years ago
You're driving down the highway and a bug spatters into your windshield. Which undergoes the greater change in momentum during t
Valentin [98]
If you are driving down the highway and a bug splatters on your windshield which undergoes the greatest change in momentum during the time of contact? momentum is the same!
8 0
2 years ago
An object is moving at a speed of 20 m/s and has a kinetic energy of 10,000 J. What is the mass of the object?
Ratling [72]

Answer:

m=100kg

Explanation:

E=1/2mv^2

10000=1/2m(20)^2

10000=1/2m(400)

10000=1/2m200

50=1/2m

100=m

3 0
3 years ago
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