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AnnZ [28]
3 years ago
10

A sled travels 15 meters down a slope inclined at 30 degrees with a horizontal. What is the horizontal displacement of the sled?

Physics
1 answer:
Arte-miy333 [17]3 years ago
6 0

Length of the slope is given as

L = 15 m

also the inclination is given as

\theta = 30 degree

now the horizontal displacement is let say "x"

now from geometry we can say

\frac{x}{L} = cos30

x = L cos30

now substitute all values in it

x = 15 * cos30

x = 12.99 m

<em>so it will displace horizontally by 12.99 m</em>

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Write a collision scenario here. If you choose your own collision, you can have neither, one, or both of the objects break. Be s
OleMash [197]

Answer:

My scenario would be A Car vs. a guard rail on a road.  You have a car that is coming down a Highway at a speed of 43 Mph Miles per hour (69.2018 Kmh)

And it hits a steel guardrail and the car smashes in at the front and the guardrail is only bent while the car has the bumper and the hood along with the headlights and windshield along with the passenger side window break.

Explanation:

This is caused by so much force reacting from one object to another but also depends on molecular density.

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Answer:

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4 0
3 years ago
Electrical power companies sell electrical energy
mixer [17]

Heat used by electric heater :

Q = m • c • ∆T

Q = (75 kg)(4200 J/kg°C)(43°C - 15°C)

Q = 8.82 × 10⁶ J

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Cost = (8.82 × 10⁶ J)/(3.6 × 10⁶ J) • ($ 0.15)

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4 0
2 years ago
A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f
Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram.  There are three forces acting on the car.  Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank.  If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

4 0
2 years ago
Jessica is running a 10K. She alternates between running and walking each kilometer. She runs at a rate of 1 kilometer every 5 m
saveliy_v [14]

Answer:

c.100 minutes

Explanation:

Total distance = 10 km

Runs for 1 km every 5 minutes

walks 1 km every 15 min

She alternates between walking and running  so,  Jessica will walk 5 km and run 5 Km

Time taken by Jessica for walking : 5 km

Time taken to walk 1 km=5 minutes

Time taken to walk 5 km

=> 5 X 5

=>25 minutes

Time taken by Jessica for Running : 5km

Time taken to run 1 km = 15 minutes

 

=> 5 X 15

=>75 minutes

Total time taken   = Time taken by Jessica for walking + Time taken by Jessica for Running

=>25 minutes +75 minutes

=> 100 minutes

8 0
3 years ago
Read 2 more answers
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