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Rus_ich [418]
3 years ago
11

Which of the following is a terrestrial habitat?A) Pond B) Garden C) Lake D) River​

Physics
1 answer:
kenny6666 [7]3 years ago
8 0

Answer:

garden

Explanation: All the other habitats are aquatic

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Plzzz help me with this<br> I’ll give brainliest
zysi [14]

Answer:

(A) By reducing friction

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2 years ago
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The combined-gas law relates which of these?
Fofino [41]
The combined-gas law relates which temperature, pressure and volume.

Temperature=T
Pressure=P
Volume=V

(P₁*V₁) / T₁=(P₂*V₂) / T₂

D. Temperature, pressuere and volume.
5 0
3 years ago
11) A sled is initially given a shove up a frictionless 35º incline. It reaches a maximum height of 2.5
Andrej [43]

Answer:

7 m/s

Explanation:

To solve this problem you must use the conservation of energy.

K1 +U1=K2+U2

That math speak for, initial kinetic energy plus initial potential energy equals final kinetic energy plus final potential energy.

The initial PE (potential energy) is 0 because it hasn't been raised in the air yet. The final KE (kinetic energy) is 0 because it isn't moving. This gives the following:

KE1= \frac{1}{2}mv^{2}}

PE2=mgh

K1=U2

\frac{1}{2} mv^{2} =mgh

Solve for v

v=\sqrt{2gh}

Input known values and you get 7 m/s.

5 0
2 years ago
How are heat and light waves produced on and in the sun?
enyata [817]
All of the electromagnetic energy radiated from the sun (and from
other stars) is the product of nuclear fusion in its core.
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2 years ago
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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
3 years ago
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