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2 years ago

What is one characteristic of informal writing?

Physics

2 answers:

7nadin3 [17]2 years ago

8 0

D. It is personal in style.

muminat2 years ago

4 0

It seems to me that the answer is D , it is personal in style

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Yasmin is testing an unknown solution to determine whether it is an acid or a base. She dips red litmus paper in the solution. I

victus00 [196]

The correct answer is D) It is not a base

7 0

2 years ago

Read 2 more answers

Question 2 (1 point)

mixer [17]

Answer:

The initial vertical velocity is zero, u = 0 m/s

Explanation:

Given;

height of the table, h = 0.55 m

horizontal distance traveled by the tennis, x = 0.12 m

Apply the following kinematic equation;

h = ut + ¹/₂gt²

where;

u is the initial vertical velocity = 0, since the tennis ball rolled off the edge of a table.

h = ¹/₂gt²

The time to fall from the vertical height is given by;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2(0.55)}{9.8} }\\\\t = 0.335 \ s$

The initial horizontal velocity of the tennis is given by;

x = vₓt

vₓ = x / t

vₓ = (0.12) / (0.335)

vₓ = 0.358 m/s

Therefore, the initial vertical velocity is zero, u = 0 m/s and initial horizontal velocity, vₓ is 0.358 m/s

3 0

3 years ago

Which of the following statements are true with respect to the law of conservation of mass?

Mazyrski [523]

Answer:

Only (I) is true

i) Mass is conserved during a chemical reaction.

8 0

2 years ago

sledge (including load) weighs 5000 N. It is pulled on level snow by a dog team exerting ahorizontal force on it. The coeﬃcient

lara31 [8.8K]

Answer:

2.5 x 10^{5} J

Explanation:

weight = 5,000 N

coefficient of friction = 0.05

distance = 1000 m

how much work is done by the dogs pulling the sledge

work done = force x coefficient of friction x distance

work done = 5000 x 0.05 x 1000 = 2.5 x 10^{5} J

6 0

3 years ago

The electric motor of a model train accelerates the train from rest to 0.685 m/s in 21.5 ms. The total mass of the train is 875

Natali [406]

Answer:

P=9.58 W

Explanation:

According to Newton's second law, and assuming friction force as zero:

$F_m=m.a\\F_m=0.875kg*a$

The acceleration is given by:

$a=\frac{\Delta v}{t}\\a=\frac{0.685m/s}{21.5*10^{-3}s}\\\\a=31.9m/s^2$

So the force exerted by the motor is:

$F_m=0.875kg*31.9m/s^2\\F_m=27.9N$

The work done by the motor is given by:

$W_m=F_m*d\\\\d=\frac{1}{2}*a*t^2\\d=\frac{1}{2}*31.9m/s^2*(21.5*10^{-3}s)^2\\\\d=7.37*10^{-3}m$

$W_m=27.9N*7.37*10^{-3}m\\W_m=0.206J$

And finally, the power is given by:

$P=\frac{W_m}{t}\\P=\frac{0.206J}{21.5*10^{-3}s}\\\\P=9.58W$

8 0

2 years ago