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Masja [62]
4 years ago
8

The zone along the southern margins of the Sahara is called the

Physics
1 answer:
Jobisdone [24]4 years ago
7 0
We have to choose the correct name for the zone along the southern margins of the Sahara. Laterite is a soil in hot and wet tropical areas. Savanna is the tropical grassland. It has tropical savanna climate. Veldt is name for the areas in the South Africa. Finally, the Sahel is the zone along the south margins of the Sahara. It has a semi-arid climate. The Arabic word "sahel" means "coast". Answer: C. Sahel. <span> </span>
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A 5 kg fish swimming at 1 m/s swallows an absentminded 500 g fish swimming toward it at a velocity that brings both fish to a ha
AlexFokin [52]

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum is defined as the product between mass and velocity of each body. And its conservation as the equality between the initial and final momentum. Mathematically described as

m_1u_1+m_2u_2 = (m_1+m_2)v_f

Here

m_1 = Mass of big fish

m_2 = Mass of small fish

v_1 = Velocity of big fish

v_2 = Velocity of small fish

v_F = Final Velocity

The big fish eats small fish and the final velocity is zero. Rearrange the equation for the initial velocity of small fish we have

m_1u_1=-m_2u_2

u_2 = -\frac{m_1u_1}{m_2}

Replacing we have,

u_2 = -\frac{(5kg)(1m/s)}{0.5kg}

u_2 = -10m/s

The negative sign indicates that the small fish is swimming in the direction opposite to that of the big fish.

Therefore the speed of the small fish is 10m/s

8 0
3 years ago
A trunk is dragged 3.0 meters across an attic floor by a force of 2.0N how much positive work is done on the trunk
Novay_Z [31]

The formula for work is

F*d

Therefore work=2.0N*3.0=6N*m

8 0
3 years ago
Las condiciones iniciales de un gas son 3000 cm3
slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

PV=nRT

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C

hence, T'=92.70°C

8 0
3 years ago
Please help me on this
QveST [7]

Answer:

i think its bulb 2

Explanation:

4 0
2 years ago
A race car has a mass of 820 kg. It starts from rest and travels 50.0m in 3.0s. The car is uniformly accelerated during the enti
Contact [7]
A=DELTAv/DELTAt=50/3
f=ma=820.50/3
7 0
3 years ago
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