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1 year ago

15

A 0. 25-kg toy car experiences a net force of 9. 0 n while slowing down to a complete stop. What is the magnitude of its acceler

ation?.

1 answer:

mezya [45]1 year ago

6 0

Answer:

a = 36 m/s

Explanation:

Given:

F = 9 N

m = 0.25 kg

a= ?

Use the equation:

F = ma (divide by m on both sides to get a alone)

a = F/m= 9/0.25 = 36 m/s

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A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de

EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/π $r^{2}$

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / π $r^{2}$ ex

E = $\frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6} } = 17.368 * 10^{9} Pa = 17.4 GPa$

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = $\frac{1}{E}$ (бy - v (бx + бz)) = $-\frac{v}{E}$ бx

= $\frac{vFx}{Epir^{2} }$ = $\frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9} } = -0.63 *10^{-6}$

Finally

ey = Δr / r

Δr = ey * r = 10 * $-0.63* 10^{-6} mm = -6.3 * 10^{-6} mm$

Δd = 2Δr = $-12.6 * 10^{-6} mm$

Explanation:

5 0

3 years ago

A solar panel measures 0.57 m by 1.3 m. In direct sunlight, the panel delivers 6.6 A at 6 V. If the intensity of sunlight is 300

zheka24 [161]

Answer:

efficiency of solar panel is 18%

Explanation:

Energy received by the sunlight is given as

$Power = intensity \times area$

here we know the dimensions of the plate as

$L = 0.57 m$

$W = 1.3 m$

now we have

$Area = (0.57)(1.3) m^2$

$A = 0.741 m^2$

now the power received by the sun light is given as

$P_1 = 0.741(300) = 222.3 W$

now the output power due to solar panel is given as

$P_{out} = V i$

$P_{out} = (6.6 A)(6 V)$

$P_{out} = 39.6 Watt$

now the efficiency is given as

$\eta = \frac{P_{out}}{P_{in}}$

$\eta = \frac{39.6}{222.3}$

$\eta = 0.18$

8 0

3 years ago

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu

dalvyx [7]

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

5 0

3 years ago

A yo-yo is made of two solid cylindrical disks, each of mass 0.055 kg and diameter 0.070 m , joined by a (concentric) thin solid

DedPeter [7]

Answer: IM 95%sure that the answer is B jus took the test got the answer right

Explanation:

6 0

3 years ago

Read 2 more answers

A man starts from rest and accelerates at 4.00 m/s2. If he covers a distance of 525 m, how long does he accelerate?

rosijanka [135]

Answer:

16.2 s

Explanation:

Given:

Δx = 525 m

v₀ = 0 m/s

a = 4.00 m/s²

Find: t

Δx = v₀ t + ½ at²

525 m = (0 m/s) t + ½ (4.00 m/s²) t²

t = 16.2 s

5 0

3 years ago