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2 years ago

Which robotic missions studied Venus?

2 answers:

6 0

The first successful flyby of Venus was performed by NASA's Mariner 2 spacecraft on 14 December 1962, following failed attempts by both the Soviet Union and the USA. The first successful landing was the Soviet Venera 4 lander, which touched down on the surface on 18 October 1967

yuradex [85]2 years ago

3 0

The first successful flyby of Venus was performed by NASA's Mariner 2 spacecraft on 14 December 1962, following failed attempts by both the Soviet Union and the USA. Hope this helps

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5. What is the velocity of a 0.5 kg ball that has a momentum of 3.00 kg m/s?

kow [346]

Answer:

6 m/s

Explanation:

p=mv

p=3 kg.m/s

m=0.5 kg

rearrange and substitute to find v:

(3)=(0.5)v

v=3/0.5

v=6 m/s

3 0

2 years ago

An old-fashioned LP record rotates at 33 1/3 RPM

Ksenya-84 [330]

Answer:

Part a) $\frac{5}{9}\ \frac{rev}{sec}$

Part b) $\frac{9}{5}\ \frac{sec}{rev}$

Explanation:

Part a) what is its frequency, in rev/s

we have that

An old-fashioned LP record rotates at 33 1/3 RPM

$33\frac{1}{3}\ \frac{rev}{min}$

Convert mixed number to an improper fraction

$33\frac{1}{3}\ \frac{rev}{min}=\frac{33*3+1}{3}=\frac{100}{3}\ \frac{rev}{min}$

Remember that

$1\ min=60\ sec$

Convert rev/min to rev/sec

$\frac{100}{3}\ \frac{rev}{min}=\frac{100}{3}(\frac{1}{60})=\frac{100}{180}\ \frac{rev}{sec}$

Simplify

$\frac{5}{9}\ \frac{rev}{sec}$

Part b) what is it period, in seconds

we know that

The period is the reciprocal of the frequency

therefore

the frequency is

$\frac{9}{5}\ \frac{sec}{rev}$

4 0

3 years ago

An electric heater is rated 1000watts. If a current of 5A pass through the heater, find the value of its resistance.

ICE Princess25 [194]

Answer: R=40 Ω

Explanation: $P=I^{2} R$

R=1000/ (25)=40

5 0

3 years ago

A fan cart initially has an acceleration of 1.6m/s/s when it's fan is directed straight backwards. If you rotate the fan by 45°,

Sholpan [36]

Answer:

A Fan Cart Initially Has An Acceleration Of 1.6m/s/s When It's Fan Is Directed Straight Backwards. If You Rotate The Fan By 45o, By What Percentage Do You Expect The Fan Cart's Thrust To Decrease? (Answer Should Be In Units Of 96)

a. 45%

b. 29%

c. 71%

d. 50%

The correct answer is d.

d. 50%

Explanation:

Fan cart acceleration = 1.6 m/s²

Thrust = 0.25×π×D²×ρ×v×Δv

where Δv = acceleration component and all factors remaining cconstant, when the fan is rotated by 45 ° the diameter changes to D₂ = sin 45 ×D

or 0.707×D. The thrust becomes 0.25×π×(0.707×D)²×ρ×v×Δv

=0.25×π×0.5×D²×ρ×v×Δv or 0.5(0.25×π×D²×ρ×v×Δv)

That is the thrust reduces by 50 %

3 0

3 years ago

A crane does 300 Joules of work in 6 seconds. How much power was used?

HACTEHA [7]

Answer: P= mad/t or P=w/t so P= 300/6= 50 W

6 0

2 years ago