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3 years ago

Which would be the most reliable source of information about the weather?

Physics

2 answers:

o-na [289]3 years ago

8 0

Answer:

C: The national weather service channel.

Explanation:

first, you can rule out A and B because they don't exactly provide exact information. D can be ruled out because it shows predicted weather while C, The national weather service channel, has 'round the clock info instead of predictions.

puteri [66]3 years ago

4 0

It is answer a a fishing magazine

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Y u y u bully me121212121

Ede4ka [16]

Answer:

wassup

Explanation:

3 0

3 years ago

Read 2 more answers

Galilee said that if you rolled a ball along a level surface it would be

lana [24]

The answer would be stay because the surface is flat so it will stay!

3 0

3 years ago

The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

5 0

3 years ago

Read 2 more answers

the base of a rectangular vessel measure 10m by 18cm. water is poured into a depth of 4cm. (a) what is the pressure on the base?

Alex787 [66]

Answer:

a) P =392.4[Pa]; b) F = 706.32[N]

Explanation:

With the input data of the problem we can calculate the area of the tank base

L = length = 10[m]

W = width = 18[cm] = 0.18[m]

A = W * L = 0.18*10

A = 1.8[m^2]

Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:

P = density * g *h

where:

density = 1000[kg/m^3]

g = gravity = 9.81[m/s^2]

h = heigth = 4[cm] = 0.04[m]

P = 1000*9.81*0.04

P = 392.4[Pa]

The force can be easily calculated knowing the relationship between pressure and force:

P = F/A

F = P*A

F = 392.4*1.8

F = 706.32[N]

4 0

3 years ago

A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because

amid [387]

Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

u2= 0m/s

v1= 10m/s

Required

The speed of the sticks

We know that from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

2000*20+1500*0=2000*10+1500*v2

40000=20000+1500v2

collect like terms

40000-20000= 1500v2

20000= 1500v2

v2= 20000/1500

v2= 13.33 m/s

Hence the velocity of the sticks is 13.33m/s

8 0

3 years ago