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Mademuasel [1]
3 years ago
13

A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen

terline at a point 1.15 m behind the front axle. Find the force exerted by the ground on each wheel.
Physics
1 answer:
NeTakaya3 years ago
4 0

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

F_1 + F_2 = W

F_1 + F_2 = (1530\times 9.81)

F_1 + F_2 = 15009.3 N

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

F_1(1.15) = F_2(2.70 - 1.15)

F_1 = 1.35 F_2

now from above equation

F_2 + 1.35F_2 = 15009.3

now we have

F_2 = 6392.85 N

now the other force is given as

F_1 = 8616.45 N

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In a nuclear fusion reaction two 2H atoms are combined to produce one 4He.
Mrrafil [7]

Answer:

a)= 0.025602u

b) = 23.848MeV

c) N = 1.546 × 10¹³

Explanation:

The reaction is

²₁H   +   ²₁H   ⇄   ⁴₂H + Q

a) The mass difference is

Δm = 2m(²₁H) - m (⁴₂H)

       = 2(2.014102u) - 4.002602u

        = 0.025602u

b) Use the Einstein mass energy relation ship

The enegy  release is the mass difference times 931.5MeV/U

E = (0.025602) (931.5)

   = 23.848MeV

c)

the number of reaction need per seconds is

N = Q/E

     = 59W/ 23.848MeV

  = \frac{59}{(23.848 \times 10^6 )(1.6 \times 10^1^9) } \\\\= 1.546 \times 10^1^3

N = 1.546 × 10¹³

5 0
3 years ago
A bionic man running at 6.5 m/sec , east is acceleration at a uniform rate of 1.5 m/sec^2 east over a displacement of 100.0 m ea
tangare [24]
Given:
u = 6.5 m/s, initial velocity 
a = 1.5 m/s², acceleration
s = 100.0 m, displacement

Let v =  the velocity attained after the 100 m displacement.
Use the formula
v² = u² + 2as
v² = (6.5 m/s)² + 2*(1.5 m/s²)*(100 m) = 342.25 (m/s)²
v = 18.5 m/s

Answer: 18.5 m/s
5 0
3 years ago
Plz i need help for the 5 problems. plz show the work!!!
Artemon [7]

Answer:

1.   3 m/s^{2}

2.   1.5 m/s^{2}

3.   3 seconds

4.   0 m/s^{2}

5.   2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

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

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = \frac {30-0}{10}=3 m/s^{2}

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

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

Substituting u=10m/s, v=22m/s and t=8 seconds

a = \frac {22-10}{8}=1.5 m/s^{2}

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

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

Substituting u=0m/s since at rest, v=15m/s and a=5 \frac {m}{s^{2}}

= \frac {15-0}{5}=3s

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

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

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = \frac {20-20}{10}=0 m/s^{2}

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

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

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 \frac {m}{s^{2}}

= \frac {0-9}{-4.1}=2.2s

8 0
3 years ago
When reading the printout from a laser printer, you are actually looking at an array of tiny dots.
solniwko [45]

Answer:

The value is y  = 3.097 * 10^{-5} \  m

Explanation:

From the question we are told that

The diameter of the pupil is d_p  =  4.2 \ mm  =  4.2 *10^{-3} \  m

The distance of the page from the eye d =  29 \  cm  =  0.29 \  m

The wavelength is \lambda  =  500 \ nm =  500 *10^{-9} \  m

The refractive index is n_r =  1.36

Generally the minimum separation of adjacent dots that can be resolved is mathematically represented as

y  = [ \frac{1.22 *  \lambda }{d_p * n_r } ]* d

         y  = [ \frac{1.22 *  500 *10^{-9} }{4.2 *10^{-3} * 1.36} ]* 0.29

         y  = 3.097 * 10^{-5} \  m

7 0
3 years ago
Which genotype has a 50% chance of being inherited
stepan [7]
Recessive genotype I hope this helps
8 0
3 years ago
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