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2 years ago

4. Inflection best relates to A. vocabulary. B. volume. C. tone. D. pace.

Physics

1 answer:

Basile [38]2 years ago

8 0

The correct answer is C) TONE

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Are you sure it’s mechanical energy? I haven’t taken a physics class in a while but mechanical energy doesn’t sound right.

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2 years ago

Một ô tô bắt đầu chuyển động thẳng nhanh dần đều,sau 10 giây đạt vận tốc 72km/h.Gia tốc của ô tô là

NeTakaya

Answer:

van

Explanation:

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2 years ago

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You are driving at 25 m/s with your cruise control on when you see a fallen tree in the road. It takes you 0.30 s to put on the

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Explanation:

Given data

velocity v= 25m/s

The time it takes to put on brake t= 0.3s

the distance covered when the brake was put on is

v=s/t

s= v*t

s= 25*0.3s

s= 7.5m

hence the distance covered is 7.5m

Also the rate of decrease in aceleration is 5m/s^2

we can also calculate the distance covered at this rate

v^2=u^2+2as

25^2= 0+2*5*s

625=10s

divide both sides by 10

s=625/10

s= 62.5m

The total distance covered between putting on the brakes and decelareation is 7.5+62.5= 70m

Given that the tree is 75m ahead, the car would not hit the tree

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2 years ago

A baseball approaches home plate at a speed of 44.0 m/s, moving horizontally just before being hit by a bat. The batter hits a p

Luda [366]

Explanation:

It is given that,

Speed of the baseball, u = 44 m/s

Speed of the baseball, v = 53 m/s

Mass of the ball, m = 145 g = 0.145 kg

Time of contact between the ball and the bat, t = 2.2 ms = 0.0022 s

$F=ma$

$F=\dfrac{mv}{t}$

$F_1=\dfrac{0.145\ kg\times 44\ m/s}{0.0022\ s}$

F₁ = 2900 N...........(1)

$F=ma$

$F=\dfrac{mv}{t}$

$F_2=\dfrac{0.145\ kg\times 53\ m/s}{0.0022\ s}$

F₂ = 3493.18 N.........(2)

In average vector form force is given by :

$F=F_1+F_2$

$F=(2900i+(-3493.18)\ N$

$F=(2900i-3493.18j)\ N$

Hence, this is the required solution.

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3 years ago

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar

horrorfan [7]

Answer:

W= 4.89 KJ

Explanation:

Lets take

temperature of hot water T₁ = 100⁰C

T₁ = 373 K

Temperature of cold ice T₂= 0⁰C

T₂ = 273 K

The latent heat of ice LH= 334 KJ

The heat rejected by the engine Q= m .LH

Q₂= 0.04 x 334

Q₂= 13.36 KJ

Heat gain by engine = Q₁

For Carnot engine

$Q_1=\dfrac{T_1}{T_2}Q_2$

$Q_1=\dfrac{373}{273}\times 13.36$

Q₁ = 18.25 KJ

The work W= Q₁ - Q₂

W= 18.25 - 13.36 KJ

W= 4.89 KJ

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2 years ago

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