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3 years ago

7

If you are driving 128.4 km/h along a straight road and you look down for 3.0s, how far do you travel during this inattentive pe

riod?

1 answer:

ser-zykov [4K]3 years ago

8 0

Answer:

107 m

Explanation:

Convert km/h to m/s:

128.4 km/h × (1000 m / km) × (1 h / 3600 s) = 35.67 m/s

Distance = rate × time

d = 35.67 m/s × 3.0 s

d = 107 m

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EleoNora [17]

Apple hits the surface with speed 16.2 m/s

The angle made by the apple velocity with normal to the incline surface is given as 20 degree

now the component of velocity which is parallel to the surface and perpendicular to the surface is given as

$v_{perpendicular} = v cos20$

$v_{parallel} = v sin20$

so here we have

$v_{parallel} = 16.2 sin20$

$v_{parallel} = 5.5 m/s$

<em>so its velocity along the incline plane will be 5.5 m/s</em>

7 0

3 years ago

A sound wave has a frequency of 425Hz. What is the period of this wave? a) 0.00235 b) 0.807 c) 425 d) 850

swat32

the answer is a) 0.00235 because 1/425=0.00235. hope I helped!

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3 years ago

To test the hypothesis that the population mean mu=3. 6, a sample size n=14 yields a sample mean 4. 007 and sample standard devi

PolarNik [594]

The P value for the given data set is 25127. For finding P value, we have to must find the Z value.

<h3>How to get the z scores?</h3>

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

The Z value is calculated as;

$Z = \dfrac{X - \mu}{\sigma})$

Z = (X - μ) / σ

Z = (4.007 - 3.6) / 0.607

Z = 0.67051

The P value for the given data set is 25127.

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2 years ago

Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary

kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

$I=mr^2$

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

$I_A=1 \times 0.5^2\ kg.m^2$

$I_A=0.25\ kg.m^2$

T= I α

0.5= 0.25 α

$\alpha = 2\ rad/s^2$

For Wheel B

m= 1 kg

d= 2 m,r=1 m

$I_B=1 \times 1^2\ kg.m^2$

$I_B=1 \ kg.m^2$

Given that angular acceleration is same for both the wheel

$\alpha = 2\ rad/s^2$

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

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3 0

3 years ago

An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 c

Aleks04 [339]

Answer:

The height of the image is, h' = 6.0 cm

The image is erect.

Explanation:

Given data,

The object distance, u = -5 cm

The focal length of convex lens, f = 10 cm

The object height, h = 3 cm

The lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{10}=\frac{1}{v}-\frac{1}{-5}$

$\frac{1}{v}=\frac{1}{10}-\frac{1}{5}$

v = -10 cm

The magnification factor of lens

$m=\frac{-10}{-5}$

m = 2

$m=\frac{h'}{h}$

$h'=h\times m$

$h'=3\times 2$

h' = 6 cm

The height of the image is, h' = 6 cm

The image is erect.

4 0

3 years ago