Ask question

Ask question

All categories

3 years ago

5

On January 22, 1943, the temperature in Spearfish, South Dakota, rose from −4°F to 45°F in just 2 minutes. What was the temperat

ure change in Celsius degrees?
27.2 C°

40.1 C°

35.4 C°

18.7 C°

1 answer:

Nimfa-mama [501]3 years ago

4 0

The answer is the first one

You might be interested in

Question:The greater the

Romashka-Z-Leto [24]

Answer:

True

Explanation:

when the object is larger , the inertia of the object is larger so its tendency to change its state of motion is reduced is reduced.

5 0

3 years ago

Seesaw unbalanced force explain

wolverine [178]

Like a seesaw, it shows that the forces aren’t equal because if it was the seesaw would stay put

3 0

3 years ago

A mass of 2 kg traveling at 3 m/s undergoes a one-dimensional elastic collision with a group of four 1kg masses that are at rest

alukav5142 [94]

2kg + 3 kg = 5 kg as result of the conclusion

6 0

3 years ago

If you live in a very cold area, you may have seen the depth of a bank of snow shrink even though temperatures remain below the

Aleks04 [339]

<span>To answer the question above, if the day sky is clear it collects radiation. If air is dry snow sublimates faster. If both cases overlap it disappears faster where ever coldest. Thank you for posting your question here. I hope my answer helps. </span>

7 0

3 years ago

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

<u>Explanation:</u>

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0

3 years ago