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3 years ago

JUST TWO QUESTIONS PLEASEEEE HELPPPPPP IM SO LOSTTTT AND I HAVE TO SUBMIT ITTT

Physics

1 answer:

Scilla [17]3 years ago

7 0

Sry,i only the answer of that question no.1

(m)=1000kg

(u)=30m/s

(f)=5000n

(v)=0m/s

(a)=f/m

= 5000/1000

Distance(s)=?

We know that,

V^2=u^2+2as

Or,0^2=30^2+2×5×s

Or,s=910m

So,the minium distance required to stop the car is 910m.

Hope,itwillhelpya!

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When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The play

saw5 [17]

Answer:

2361.6N

Explanation:

Mass of player = 82kg

Velocity = 1.2m/s

Kinetic energy of player:

= 1/2mv²

= 1/2*82*1.2²

= 41x1.44

= 59.04J

Final kinetic energy = 0

Change in kinetic energy

|∆k| = |0-59.04|

= 59.04

Workdone by the feet = fd

d = 0.025

Fd = 59.04

F = 59.04/0.025

= 2361.6N

This is his average force.

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3 years ago

Why are adaptations important to organisms?

Serga [27]

Adaptations can help organisms reproduce and continue living in different conditions

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4 years ago

If one bullet is dropped from a certain height and another is fired from a gun horizontally from the same height which one will

Zinaida [17]

If the ground is flat, and both bullets are released at the same time from the same height, then they both hit the ground at the same time.

The horizontal motion of the one from the gun has no effect on its vertical motion.

7 0

4 years ago

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

liq [111]

Answer:

The speed of water must be expelled at 6.06 m/s

Explanation:

Neglecting any drag effects of the surrounding water we can assume the linear momentum in this case is conserves, that is, the total initial momentum of the octopus and the water kept in it cavity should be equal to the total final linear momentum. That's known as conservation of momentum, mathematically expressed as:

$p_f=p_i$

with Pi the total initial momentum and Pf the final total momentum. The total momentum is the sum of the momentums of the individual objects, in our case the octopus and the mass of water that will be expelled:

$p_{of}+p_{wf}=p_{oi}+p_{wi}$

with Po the momentum of the octopus and Pw the momentum of expelled water. Linear momentum is defined as mass times velocity:

$m_o*v_{of}+m_w*v_{wf}=m_o*v_{oi}+m_w*v_{wi}$

Note that initially the octopus has the water in its cavity and both are at rest before it sees the predator so $v_{oi}=v_{wi} = 0\frac{m}{s}$ :

$m_o*v_{of}+m_w*v_{wf}=0$

We should find the final velocity of water if the final velocity of the octopus is 2.70 m/s, solving for $v_{wf}$ :

$v_{wf}=-\frac{m_o*v_{of}}{m_w}=-\frac{(6.00-1.85)*(2.70)}{1.85}$

$v_{wf}=-6.06\frac{m}{s}$

The minus sign indicates the velocity of the water is opposite the velocity of the octopus.

3 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

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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