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RideAnS [48]
3 years ago
9

Suppose a treadmill has an average acceleration of 4.7x10^-3 m/s. a)how much does its speed change after 5min? b)if the treadmil

l's speed 1.7 m/s, what will its final speed be?
Physics
1 answer:
Mars2501 [29]3 years ago
8 0

Answer:

a)Change in the speed  = 1.41 m/s

b)The final speed will be 3.11 m/s

Explanation:

Given that

Acceleration ,a= 4.7 x 10⁻³ m/s²

a)

We know that

v= u + a t

v=final speed ,u=initial speed

t= time ,a= acceleration

Change in the speed

v- u = a t

t= 5 min  = 5 x 60 s = 300 s

v- u = 4.7 x 10⁻³ x 5 x 60 m/s

v-u = 1.41 m/s

Change in the speed  = 1.41 m/s

b)

Given that

u= 1.7 m/s

v-u = 1.41 m/s

v= 1.7 + 1.41 m/s

v=3.11 m/s

The final speed will be 3.11 m/s

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I'd say b, precise, here.
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3 years ago
Read 2 more answers
He starter motor of a car engine draws a current of 170 AA from the battery. The copper wire to the motor is 6.00 mmmm in diamet
sweet-ann [11.9K]

Answer:

129.2 C

0.33758239177 mm

Explanation:

n = Number density = 8.46\times 10^{28}\ electrons/m^3

i = Current = 170 A

t = Time taken = 0.76 s

d = Diameter = 6 mm

Charge is given by

q=it\\\Rightarrow q=170\times 0.76\\\Rightarrow q=129.2\ C

The charge passing throught the motor is 129.2 C

Current density

J=\dfrac{i}{A}\\\Rightarrow J=\dfrac{170}{\dfrac{\pi}{4}\times (6\times 10^{-3})^2}\\\Rightarrow J=6012520.07236\ A/m^2

Drift velocity is given by

v_d=\dfrac{J}{ne}\\\Rightarrow v_d=\dfrac{6012520.07236}{8.46\times 10^{28}\times 1.6\times 10^{-19}}\\\Rightarrow v_d=0.000444187357592\ m/s

Distance traveled

s=v_dt\\\Rightarrow s=0.000444187357592\times 0.76\\\Rightarrow s=0.00033758239177\ m=0.33758239177\ mm

The electron traveled 0.33758239177 mm

8 0
3 years ago
Why is water And effective solvent for many different types of compounds?
Artyom0805 [142]

B. it contains polar covalent bonds

6 0
3 years ago
6) The temperature of 10 kg of a substance rises by 55°C when heated. Calculate the
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Answer:

121

Explanation:

8 0
3 years ago
4. A bullet of mass 30 g is fired from a rifle of mass 5kg at a speed of 259m/s. 
Ostrovityanka [42]

Answer:

Rifle Momentum=7.77kg*m/s v'= 1.554 m/s

Explanation:

a) m1v1 + m2v2 = m1v1' + m2v2'

0+0 = 0.03*259 + P(rifle momentum)

solve for P

p= 7.77kg*m/s

b) 7.77= 5*v'

v'= 1.554 m/s

8 0
3 years ago
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