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slava [35]
3 years ago
12

Archer shoots his arrow towards a target at a distance of 90 m and hits bullseye . Calculate the acceleration and time taken by

the arrow to hit the target with a velocity of 90.0 m/s
Physics
1 answer:
goldfiish [28.3K]3 years ago
3 0

Answer:

acceleration = 45 m/s²

Time = 20 seconds

Explanation:

We are given;

Distance; s = 90m

Final velocity;v = 90 m/s

Since shot from rest, initial velocity;u = 0 m/s

So, to find the acceleration, we can use Newton's 2nd law of motion which is;

v² = u² + 2as

Where a is acceleration

Plugging in the values;

90² = 0² + 2a(90)

8100 = 180a

a = 8100/180

a = 45 m/s²

Now, for the time, we will use Newton's 1st law of motion which is;

v = u + at

Where t is time;

So;

90 = 0 + 4.5t

90 = 4.5t

t = 90/4.5

t = 20 seconds

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labwork [276]
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5 0
2 years ago
A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 20.1 N; when it is completely i
Paha777 [63]

Answer:

A) V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}

B) d = 4181.49 kg/m^{3} = 4.18 g/cm^{3}

Explanation:

A) Using the Archimedes' force we can find the weight of water displaced:

W_{d} = W_{a} - W_{w}

Where:

W_{a}: is the weight of the block in the air = 20.1 N

W_{w}: is the weight of the block in the water = 15.3 N

W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N

Now, the mass of the water displaced is:

m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg

The volume of the block can be found using the mass of water displaced and the density of the water:

V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}

B) The density of the block can be found as follows:

d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3}

I hope it helps you!            

6 0
3 years ago
An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and
Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is F_{ma}=BqV---(i)

The given formula to find the electric force is F_{el}=qE---(ii)

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

Bqv=qE\\\\v=\frac{E}{B}

v=\frac{187500}{0.125} \\\\v=15\times10^5m/s

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

F_{ce}=\frac{mv^2}{r}---(iii)

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br}  \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg

Therefore,  the particle's charge-to-mass ratio is 958.1\times10^5C/kg

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

8 0
3 years ago
A person who studies physics is known as
Leto [7]
Physicists. Welcome :)
4 0
3 years ago
Read 2 more answers
The driver of a car slams on the brakes when he sees a tree blocking the road. the car slows uniformly with acceleration of -5.9
Hatshy [7]
Let u =  the speed of the car at the instant when braking begins.

The braking distance is s = 62.3 m, the acceleration is a = -5.9 m/s², and the braking duration is t = 4.15 s.

Use the formula s = ut + (1/2)at² to obtain
(u m/s)*(4.15 s) + 0.5*(-5.9 m/s²)*(4.5 s)² = (62.3 m)
4.15u = 62.3 + 50.8064 = 113.1064
      u = 27.2546 m/s

Let v m/s be the speed with which the car strikes the tree.
Then
v = 27.2546 - 5.9*4.15
   = 2.7696 m/s

Answer: 2.77 m/s (nearest hundredth)

4 0
3 years ago
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