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kondor19780726 [428]

3 years ago

13

A car travels 85 km in the first half hour of a trip. The car continues to travel for 2 more hours and travels 200 km. What was

the average speed of the car for the trip?

1 answer:

viva [34]3 years ago

3 0

Answer:

50

Explanation:

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in a billiards game, the white cue ball hits the black ball, moving the black ball to the right while the cue ball moves to the

Gemiola [76]

Answer:

The black ball moving to the right after being struck by the white ball

Explanation:

4 0

3 years ago

At its closest approach, a moon comes within 200,000 km of the planet it orbits. At that point, the moon is 300,000 km from the

krok68 [10]

Answer:

s₁ = 240,000 km

Explanation:

The distance between both the focuses f₁ and f₂ will be the sum of distances of the moon from each focus at a given point. Therefore,

s = s₁ + s₂

where,

s = total distance between the focuses = ?

s₁ = distance between f1 and moon = 200,000 km

s₂ = distance between f₂ and moon = 300,000 km

Therefore,

s = 200,000 km + 300,000 km

s = 500,000 km

Now, when the distance from f₂ becomes 260,000 km, then the distance from f₁(planet) will become:

s = s₁ + s₂

500,000 km = s₁ + 260,000 km

s₁ = 500,000 km - 260,000 km

<u>s₁ = 240,000 km</u>

5 0

3 years ago

The coefficients of friction between the 20-kg crate and the inclined surface are µ,8 = 0.24 and J.lk = 0.22. If the crate start

Yanka [14]

Answer:5.60 m/s

Explanation:

Given

Coefficient of static friction $\mu _s=0.24$

Coefficient of kinetic friction $\mu _k=0.22$

mass of crate $m=20\ kg$

Force applied $F=200\ N$

maximum static Friction $F_s=\mu _sN$

$N=mg$

$F_s=0.24\times 20\times 9.8$

$F_s=47.04\ N$

thus applied force is greater than Static friction therefore kinetic friction will come into play

$F_k=\mu _kN$

$F_k=0.22\times 20\times 9.8=43.12\ N$

net Force on crate $F-F_k=ma$

$a=\frac{200-43.12}{20}=7.84\ m/s^2$

Magnitude of velocity can be obtained by using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

here initial velocity is zero as crate start from rest

$v^2-0=2\times 7.84\times 2$

$v=\sqrt{31.37}$

$v=5.60\ m/s$

7 0

4 years ago

What is the speed of a wave frequency of 300hz and a wavelength of 25M

tia_tia [17]

Answer:

7500 m/s

Explanation:

We can use the equation velocity of a wave equals wavelength times frequency. Therefore, v = wavelength*f = (25 m)(300 Hz) = m/s7,500

8 0

3 years ago

A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k

Luba_88 [7]

Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

$\theta = 40$ °

$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

$F_{fri} =$ 0.22 × 29.3

$F_{fri} =$ 6.446 N

(b). Block acceleration is given by

$F_{net} = F - F_{fri}$

F = 30 N

$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

23.554 = 3.9 × a

a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

8 0

3 years ago