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3 years ago

I need help for the first 3 plz.

Physics

2 answers:

madam [21]3 years ago

7 0

1) hypothesis
2) data
3) method

I think these are correct.

Sonbull [250]3 years ago

3 0

It's hypothesis then its data and then is method

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At one particular moment, a 19.0 kg toboggan is moving over a horizontal surface of snow at 4.00 m/s. After 7.00 s have elapsed,

Lina20 [59]

Answer:

10.86 N

Explanation:

Let the average frictional force acting on the toboggan be 'f' N.

Given:

Mass of toboggan (m) = 19.0 kg

Initial velocity (u) = 4.00 m/s

Final velocity (v) = 0 m/s

Time for which friction acts (Δt) = 7.00 s

Now, change in momentum is given as:

$\Delta p =Final\ momentum-Initial\ momentum\\\\\Delta p=mv-mu\\\\\Delta p=19.0\ kg(0-4.00)\ m/s\\\\\Delta p=-76.00\ Ns$

Now, we know that, change in momentum is equal to the impulse acting on the body. So,

Impulse is, $J=\Delta p=-76.00\ Ns$

Now, we know that, impulse is also given as the product of average force and the time interval for which it acts. So,

$J=f\times \Delta t$

Rewriting the above equation in terms of 'f', we get:

$f=\dfrac{J}{\Delta t}$

Plug in the given values and solve for 'f'. This gives,

$f=\frac{-76.00\ Ns}{7.00\ s}\\\\f=-10.86\ N$

Therefore, the magnitude of frictional force is $|f|=|-10.86\ N|=10.86\ N$

3 0

3 years ago

Does someone know how to do math with that equation

mestny [16]

Answer:

f = 5 cm

Explanation:

using the thin lens equation, given as follows:

$\frac{1}{f} = \frac{1}{d_{o}}+\frac{1}{d_{i}}\\$

where,

f = focal length = ?

do = the distance of object from lens = 20 cm

di = the distance of image from lens = 6.6667 cm

Therefore,

$\frac{1}{f} = \frac{1}{20\ cm}+\frac{1}{6.6667\ cm}\\\\\frac{1}{f} = 0.199999\ cm^{-1}\\\\f = \frac{1}{0.199999\ cm^{-1}}\\\\$

8 0

3 years ago

Greg’s mom baked 24 brownies for the bake sale. Vinnie’s mom

AysviL [449]

(24 + 36) x 2 = 120

simple math. you're welcome.

"college student"

4 0

3 years ago

Read 2 more answers

Pls help will mark Brainliest

BartSMP [9]

Answer: 100 units

Explanation:

8 0

3 years ago

Monochromatic light of wavelength 687 nm is incident on a narrow slit. On a screen 1.65 m away, the distance between the second

Sophie [7]

Answer:

a ) 1.267 radian

b ) 1.084 10⁻³ mm

Explanation:

Distance of screen D = 1.65 m

Width of slit d = ?

Wave length of light λ = 687 nm.

Distance of second minimum fro centre y = 2.09 cm

Angle of diffraction = y / D

= 2.09 /1.65

= 1.267. radian

Angle of diffraction of second minimum

= 2 λ / d

so 2 λ / d = 1.267

d = 2 λ / 1.267 = (2 x 687 ) /1.267 nm

=1084.45 nm = 1.084 x 10⁻³ mm.

3 0

3 years ago