Please help oooooooo

You leave on a 450 miles trip in order to attend a meeting that will start 10.8 hours after you begin your trip. Along the way y

Lorico [155]

Answer:2.6 h

Explanation:

Given

Total Trip distance=450 miles

Meeting starts after 10.8 hours

safe Fastest speed is 55 mi/h

so if he drives all the to the meeting with max speed then it takes $=\frac{450}{55}=8.181 h$

and total allowable time is 10.8

Therefore longest time he can spend over dinner is $10.8-8.181 \approx 2.6 hours$

5 0

3 years ago

calculate the efficiency of a light bulb that had an input of 400 j transfers 100j as a light and 300j as heat.

beks73 [17]

Efficiency = useful energy out / total energy in x 100
= 100/400 x 100
=0.25 x 100
= 25%

25%

6 0

2 years ago

Find the average speed of a car that travels 45km in 2hours 15minutes

Helga [31]

Answer:

20 km/h

Explanation:

45 km ÷ 2.25 hours (15 mins is 0.25 hours)

= 20

20 km/h

3 0

2 years ago

car is moving at 40 m/s. At 10 meters the driver spots a deer on the road and instantly steps on the brakes. If the car is 400 k

Mice21 [21]

Answer:

32000 N

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 40 m/s

Distance (s) = 10 m

Final velocity (v) = 0 m/s

Mass (m) of car = 400 Kg

Force (F) =?

Next, we shall determine the acceleration of the the car. This can be obtained as follow:

Initial velocity (u) = 40 m/s

Distance (s) = 10 m

Final velocity (v) = 0 m/s

Acceleration (a) =?

v² = u² + 2as

0² = 40² + (2 × a × 10)

0 = 1600 + 20a

Collect like terms

0 – 1600 = 20a

–1600 = 20a

Divide both side by –1600

a = –1600 / 20

a = –80 m/s²

The negative sign indicate that the car is decelerating i.e coming to rest.

Finally, we shall determine the force needed to stop the car. This can be obtained as follow:

Mass (m) of car = 400 Kg

Acceleration (a) = –80 m/s²

Force (F) =?

F = ma

F = 400 × –80

F = – 32000 N

NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.

7 0

2 years ago

A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re

valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude w, pointing downward)

• the normal force of the incline on the box (mag. n, pointing upward perpendicular to the incline)

• friction (mag. f, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ F = -f + w sin(35°) = m a

• net perpendicular force:

∑ F = n - w cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

n = w cos(35°)

n = (15 kg) (9.8 m/s²) cos(35°)

n ≈ 120 N

Solve for the mag. of friction:

f = µ n

f = 0.25 (120 N)

f ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) a

a ≈ (54.3157 N) / (15 kg)

a ≈ 3.6 m/s²

Now solve for the block's speed v given that it starts at rest, with v₀ = 0, and slides down the incline a distance of ∆x = 3 m:

v² - v₀² = 2 a ∆x

v² = 2 (3.6 m/s²) (3 m)

v = √(21.7263 m²/s²)

v ≈ 4.7 m/s

4 0

2 years ago