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ivanzaharov [21]
3 years ago
11

A dragster is traveling at a constant rate of 45 m/s. How fast is it going when it has traveled 325 m?

Physics
1 answer:
Sergeeva-Olga [200]3 years ago
6 0

The speed of the dragster after travelling 325 m is still 45 m/s

Explanation:

In this problem, we are told that the dragster is travelling at constant rate, therefore its speed is constant and it is

v = 45 m/s

Therefore, its speed will remain constant during the entire motion: therefore, after having travelled for 325 m, its speed will still be 45 m/s.

We can also calculate the time taken for the dragster to cover the 325 m. In fact, the speed in a uniform motion is given by

v=\frac{d}{t}

where:

v is the speed

d is the distance covered

t is the time elapsed

In this case, we have:

v = 45 m/s

d = 325 m

Solving for t, we find:

t=\frac{d}{v}=\frac{325}{45}=7.2 s

Learn more about speed:

brainly.com/question/8893949

#LearnwithBrainly

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frez [133]
5.91(approx) seconds just divide velocity by acceleration
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For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

4 0
3 years ago
Read 2 more answers
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.70 m/s. Her husband Bruce suddenly realiz
Katen [24]

Answer:

The velocity is v  =  6.66 \  m/s

Henrietta is at distance s=  18.17 \  m from the under the window

Explanation:

From the question we are told that

The speed of Henrietta is v=  2.70 \ m/s

The height of the window from the ground is h  =  36.5 \  m

Generally the time taken for the lunch to reach the ground assuming it fell directly under the window is

t  =  \sqrt{\frac{2 *  h }{g} }

=> t  =  \sqrt{\frac{2 *  36.5 }{9,8} }

=> t  =  2.73 \  s

Generally the time taken for the lunch to reach Henrietta is mathematically represented as

T =  t +  t_1

Here t_1 is the time duration that elapsed after Henrietta has passed below the window the value is given as 4 s

Now

T = 2.73  +  4

=> T = 6.73 \  s

Generally the distance covered by Henrietta before catching her lunch is

s=  v  *  T

=> s=  2.70  * 6.73

=> s=  18.17 \  m

Generally the speed with which Bruce threw her lunch is mathematically represented as

v  =  \frac{18.17}{2.73}

v  =  6.66 \  m/s

4 0
3 years ago
O be useful in science, a hypothesis must be *
shepuryov [24]

Answer:

the answer is c testable

3 0
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Daniel [21]

Answer:

D. are brought from the mantle to the surface in magma that hardens into komatiite.

Explanation:

Diamond :

 It is the hardest form of carbon.The atomic atoms arrange in the cubic crystal structure and this is known as diamond cubic.Another form of the diamond at room temperature is graphite.This is used for making jewelry.This is also used in the cutting process because it has high strength.

Therefore the correct option for the diamond is D.

4 0
3 years ago
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