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Nastasia [14]
3 years ago
7

10. A 25 kg apple cart is being pushed with an applied force of 115 N. The coefficient of friction between the ground and the ca

rt is 0.35. What is the acceleration of the cart?
Physics
1 answer:
ziro4ka [17]3 years ago
8 0

Answer:

1.1 m/s²

Explanation:

From the question,

F -mgμ = ma.................... Equation 1

Where F = applied force, m = mass of the apple cart, g = acceleration due to gravity, μ =  coefficient of friction., a = acceleration of the apple cart.

Given: F = 115 N, m = 25 kg,  μ  = 0.35

Constant: g = 10 m/s²

Substitute these values into equation 2

115-(25×10×0.35) = 25×a

115-87.5 = 25a

25a = 27.5

a = 27.5/25

a = 1.1 m/s²

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Answer:

A. 0.204203928 m

b. 0.1443939822

Explanation:

Givens:

Volume (V) = 160 m^3

velocity 1 (v1) = 3.00 m/s

velocity 2 (v2) = 6.00 m/s

Time (t) = 21.3 minutes = 1278 seconds

Side length = ?

V = A*v*t

A = s^2

V = s^2*v*t

s= \sqrt{\frac{V}{v*t} }

A.

s = \sqrt{\frac{V}{v*t} } = \sqrt{\frac{160}{3.00*1278} } = 0.204203928 m

B.

s = \sqrt{\frac{V}{v*t} } = \sqrt{\frac{160}{6.00*1278} } = 0.1443939822 m

I hope this helped!! Have a good day.

7 0
3 years ago
3) Find the missing parts of the triangle (angle and/or sides).<br>​
oee [108]

Answer:

44.4°

Explanation:

Use SOH-CAH-TOA.

Sine = Opposite / Hypotenuse

Cosine = Adjacent / Hypotenuse

Tangent = Opposite / Adjacent

You're given an unknown angle, the adjacent side to that angle, and the hypotenuse.  So use cosine.

Cosine = Adjacent / Hypotenuse

cos A = 10 / 14

A = cos⁻¹(5/7)

A ≈ 44.4°

8 0
3 years ago
What is the mechanical advantage supplied by a 1000 N object lifted using 250 N of force
Ivenika [448]

Answer:

Mechanical advantage=4

Explanation:

Load=1000N

Effort=250N

mechanical advantage=load/effort

Mechanical advantage=1000/250

Mechanical advantage=4

4 0
4 years ago
White light (light containing all frequencies of light in the visible spectrum) passes through a diffraction grating with 1300 l
lara31 [8.8K]

Answer:

The distance between the 1st order blue fringe and the 1st order red fringe is 6.5 cm.

Explanation:

Given that,

Diffraction grating =1300 lines/cm

Distance of screen = 2 m

Order number = 1

We need to calculate the distance between the 1st order blue fringe and the 1st order red fringe

Using formula of distance

For first order red fringe

d\sin\theta=m\lambda

d\times\dfrac{y}{l}=m\times\lambda

y_{r}=\dfrac{m\times\lambda l}{d}

y_{r}=\dfrac{1\times700\times10^{-9}\times2}{\dfrac{1}{1300}\times10^{-2}}

y_{r}=0.182\ m

For first order blue fringe

y_{b}=\dfrac{1\times450\times10^{-9}\times2}{\dfrac{1}{1300}\times10^{-2}}

y_{b}=0.117\ m

We need to calculate the distance between blue fringe and red fringe

\Delta y=y_{r}-y_{b}

\Delta y=0.182-0.117

\Delta y=0.065\ m=6.5 cm

Hence, The distance between the 1st order blue fringe and the 1st order red fringe is 6.5 cm.

7 0
3 years ago
Give me one example of a sport that requires someone to have good reaction time?
DaniilM [7]

Answer:

Baseball

Explanation:

Baseball because the ball can come at you at high speeds and you need to react quickly.

4 0
3 years ago
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