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Suppose that you are standing on a train accelerating at 0.20g. What minimum coefficient of static friction must exist between y

Ilia_Sergeevich [38]

Acceleration = (0.2 x g) = 1.96m/sec^2.
Accelerating force on 1kg. = (ma) = 1.96N. 
1kg. has a weight (normal force) of 9.8N. 
Coefficient µ = 1.96/9.8 = 0.2 minimum. 

Coefficient is a ratio, so holds true for any value of mass to find accelerating force acting. 
e.g. 75kg = (75 x g) = 735N. 
Accelerating force = (735 x 0.2) = 147N

5 0

4 years ago

Read 2 more answers

Match the correct percentage of water to the correct phrase

Nesterboy [21]

Answer:

b

Explanation:

get a life <3

5 0

3 years ago

A 35 kg child goes down a playground that is inclined at an angle of 27.5 degrees above the horizontal. Find the acceleration if

tigry1 [53]

Answer:

4.16m/s²

Explanation:

According to Newtons second law;

$\sum Fx = ma_x\\Fm - Ff = ma_x\\W sin\theta - \mu R cos \theta = ma_x\\mg sin\theta - \mu mg cos\theta = ma_x\\$

Fm is the moving force

$\mu$ is the coefficient of kinetic friction between the child and the slide

m is the mass

g is the acceleration due to gravity

a is the acceleration of the child

Substitute the given values and get the acceleration as shown;

35(9.8)sin27.5 - 0.415(35)(cos27.5) = 35a

158.38-12.88 = 35a

145.49 = 35a

a = 145.49/35

a = 4.16m/s²

Hence the acceleration of the body is 4.16m/s²

4 0

3 years ago

An α-particle has a charge of +2e and a mass of 6.64 × 10-27 kg. It is accelerated from rest through a potential difference that

zzz [600]

Answer with Explanation:

We are given that

Charge on alpha particle=q=2 e= $2\times 1.6\times 10^{-19} C$

1 e= $1.6\times 10^{-19} C$

Mass of alpha particle=m= $6.64\times 10^{-27} kg$

Potential difference,V= $1.97\times 10^6 V$

Magnetic field,B=3.49 T

a.Speed of alpha particle=v= $\sqrt{\frac{2 qV}{m}}$

By using the formula

$v=\sqrt{\frac{2\times 2\times 1.6\times 10^{-19}\times 1.97\times 10^6}{6.64\times 10^{-27}}$

$v=1.38\times 10^7 m/s$

b.Magnetic force,F $=qvB=2\times 1.6\times 10^{-19}\times 1.38\times 10^7\times 3.49=1.5\times 10^{-11} N$

F= $1.5\times 10^{-11} N$

c.Radius of circular path, r= $\frac{mv^2}{F}$

$r= \frac{6.64\times 10^{-27}\times (1.38\times 10^7)^2}{1.5\times 10^{-11}}$

r=0.084 m

5 0

3 years ago

A certain car's drive-train produces a force of 5300 N as it accelerates from 0

lakkis [162]

The power is $7.1\cdot 10^4 W$

Explanation:

First of all, we need to find the acceleration of the car, which is given by

$a=\frac{v-u}{t}$

where

v = 60 mph = 26.8 m/s is the final velocity

u = 0 is the initial velocity

t = 10.0 s is the time

Substituting,

$a=\frac{26.8-0}{10.0}=2.68 m/s^2$

Now we can find the mass of the car by using Newton's second law:

$F=ma$

where

F = 5300 N is the force applied

m is the mass

$a=2.68 m/s^2$ is the acceleration

Solving for m,

$m=\frac{F}{a}=\frac{5300}{2.68}=1978 kg$

Now we can use the work-energy theorem, which states that the work done is equal to the change in kinetic energy of the car, to find the work:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$

And substituting,

$W=\frac{1}{2}(1978)(26.8)^2-0=7.10\cdot 10^5 J$

Finally, we can find the power output of the car:

$P=\frac{W}{t}$

where

W is the work

t = 10.0 s is the time elapsed

Substituting,

$P=\frac{7.1\cdot 10^5}{10.0}=7.1\cdot 10^4 W$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

4 0

4 years ago