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3 years ago

Calculate the speed of an 8.0 × 104 kg airliner with a kinetic energy of 1.1 × 109 J.

Physics

1 answer:

Mazyrski [523]3 years ago

5 0

KE = ½mv²
<span>1.1 x 10^9 = ½ x 8 x 10^4 v²
v² = 2.75 x 10^4
v ≈ 165.8 m/s</span>

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List two industrial uses for nonmetallic mineral resources

Anni [7]

Answer:

To make useful chemicals and abrasives

Explanation:

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3 years ago

A sprinter set a high school record in track and field, running 200.0 m in 20.6 s . what is the average speed of the sprinter in

Paraphin [41]

Answer : The average speed of the sprinter is, 34.95 Km/hr

Solution :

Average velocity : It is defined as the distance traveled by the time taken.

Formula used for average velocity :

$v_{av}=\frac{d}{t}$

where,

$v_{av}$ = average velocity

d = distance traveled = 200 m

t = time taken = 20.6 s

Now put all the given values in the above formula, we get the average velocity of the sprinter.

$v_{av}=\frac{200m}{20.6s}\times \frac{3600}{1000}=34.95Km/hr$

conversion :

(1 Km = 1000m)

(1 hr = 3600 s)

Therefore, the average speed of the sprinter is, 34.95 Km/hr

8 0

3 years ago

Read 2 more answers

A ball is released from the top of a tower of height h meters. it takes T seconds to reach the ground . what is the position of

alekssr [168]

Answer:the

8/9 h

Explanation:

Height = 1/2 a T^2 now change to T/3

now height = 1/2 a (T/3)^2 =<u> 1/9</u> 1/2 a T^2 <===== it is 1/9 of the way down or 8/9 h

3 0

2 years ago

As a physics instructor hurries to the bus stop, her bus passes her, stops ahead, and begins loading passengers. She runs at 6.0

Alika [10]

velocity of the physics instructor with respect to bus

$v = 6 m/s$

acceleration of the bus is given as

$a = 2 m/s^2$

acceleration of instructor with respect to bus is given as

$a = -2 m/s^2$

now the maximum distance that instructor will move with respect to bus is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - 6^2 = 2(-2)(d)$

$-36 = - 4 d$

$d = 9 m$

so the position of the instructor with respect to door is exceed by

$\delta x = 9 - 6 = 3 m$

so it will be moved maximum by 3 m distance

7 0

4 years ago

A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the

Goshia [24]

To solve this problem we will use the concepts related to power, defined as the amount of energy applied over a period of time.

The energy in this case is the accumulated in the form of potential energy, over a period of time. Thus we will have that the mathematical expression of the power can be expressed as

$P = \frac{E}{t}$

Here,

E = Energy

t = time

As the energy is equal to the potential Energy we have tat

$P = \frac{mgh}{t}$

The weight (mg) of the man is 700N, the height (h) is 8m and the time is 10s, then:

$P = \frac{700*8}{10}$

$P = 560W$

Therefore the correct answer is A.

8 0

4 years ago