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ryzh [129]
3 years ago
5

Which element is most likely to bend without breaking?

Physics
2 answers:
Pepsi [2]3 years ago
5 0

Answer:

C. cobalt (Co)

Explanation:

For people on ed2020

KengaRu [80]3 years ago
4 0

Answer:

Cobalt (Co)

Explanation:

Cobalt (Co) is a metal. Most metals are ductile and malleable, which means that in their solid form they can be bend and transformed into thin sheets and wires without braking.

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lys-0071 [83]
The net force is 12 N to the left.
4 0
2 years ago
Read 2 more answers
2. physical quantities which do not depend on any physical quantities for their measurements are known as​
seropon [69]

Answer:

Fundamental quantities

5 0
3 years ago
A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th
ANEK [815]

Answer:\dfrac{\pi}{2} ms^{-1}

Explanation:

Let T be the time required to make one revolution.

Let r be the radius of the circular path.

Let d be the distance travelled by ball in one revolution.

As we know,the distance travelled in one revolution is the circumference of the circle.

So,d=2\pi r

Given,d=0.5m\\T=2sec

d=2\times \pi \times 0.5=\pi m

Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.

Let s be the speed of the ball.

s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}

So,the speed of the ball is \frac{\pi }{2}ms^{-1}

5 0
3 years ago
A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what s
3241004551 [841]

Answer:

The speed the bat is gaining on its prey is 0.03m/s

Explanation:

Given;

speed of the bat, v₀ = 3.7 m/s

frequency of the bat, F₀ = 36 kHz

frequency of the source, Fs = 36.79

This is relative motion between a source of the sound and the observer.  The phenomenon is known as Doppler effect.

Apply the following equation to determine the speed of the insect which is the source;

F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\  340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s

The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s

Therefore, the speed the bat is gaining on its prey is 0.03m/s

8 0
3 years ago
Anna Litical analyzes the force between a planet and its moon, varying the mass of
Travka [436]

Answer:

Trial 1 is the largest, trial 3 is the smallest

Explanation:

Given:

<em>Trial 1</em>

M₁ = 6·10²² kg

d₁ = 3 500 km = 3.5·10⁶ м

<em>Trial  2</em>

M₂ = 6·10²² kg

d₂ = 7 000 km = 7·10⁶ м

<em>Trial  3</em>

M₃ = 3·10²² kg

d₃ = 7 000 km = 7·10⁶ м

___________

F - ?

Gravitational force:

F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m  (N)

F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m  (N)

F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m  (N)

Trial 1 is the largest, trial 3 is the smallest

5 0
1 year ago
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