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3 years ago

PLEASE HELP!!

Physics

1 answer:

BartSMP [9]3 years ago

4 0

Answer:A) 19.8 seconds

B) 467.5 metres

Explanation: 100km/hr = 27.78m/s

So, it covers 27.78 metres in one second

Take the time it would cover 550m as x

Therefore 27.78m = 1second

550m = x

Cross multiplying

27.78x = 550m

Divide both sides by the coefficient of x

x = 550÷27.78

x = 19.8seconds

B) distance = speed × time

Speed = 85km/hr = 23.61m/s

Time = 19.8seconds

Therefore,

Distance = 23.61 × 19.8

Distance = 467.5m

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A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to

gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

$v=at+v_o$ where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

$v=at+v_o$

$(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})$

$1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t$

$\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}$

$10\text{ [s]}=t$

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

Verification:

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is $a=\dfrac{100 [\frac{m}{s}]}{1[s]}$ , the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

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2 years ago

Which group in the periodic table is known as salt farmers?

adell [148]

Which group in the periodic table is known as salt formers?

The correct option is the last one: Halogen family.

You can find the halogen or "salt formers" in the group 17 of the periodic table. These are:

- Fluorine.
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- Bromine.
- Iodine.
- Astatine.

All of them are non-metallic elements and they have 7 electrons.

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3 years ago

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mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan

Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, $v_e$ , for the planet is $v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }$

Where;

$v_e$ = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, $M_E$

r = The radius of the planet = 3 × The radius of Earth, $R_E$

The escape velocity for Earth, $v_e_E$ , is therefore given as follows;

$v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }$

$\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } = \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E$

$v_e$ = 16 × $v_e_E$

Given that the escape velocity for Earth, $v_e_E$ ≈ 11,186 m/s, we have;

The escape velocity on the planet = $v_e$ ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

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3 years ago

A heavy rope, 80 ft long and weighing 32 lbs, hangs over the edge of a building 100 ft high. how much work w is done in pulling

allochka39001 [22]

The first thing you should know for this case is that work is defined as the product of force by the distance traveled in the direction of force.
We have then:
W = Fd
The distance varies, so we must integrate:
from 0 to 20:
W = ∫F (x) dx
W = ∫32xdx
W = 32∫xdx
W = 32 (x ^ 2/2) = (16) (20 ^ 2) = 6400 ft * lbs
answer:
6400 ft * lbs is work done pulling the rope up 20 ft

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3 years ago

A brand new corvette can go from 0 to 85 miles per hour in 4.8 seconds.

klasskru [66]

Answer:

(A) 7.9 m/s^{2}

(B) 19 m/s

(C) 91 m

Explanation:

initial velocity (U) = 0 mph = 0 m/s

final velocity (V) = 85 mph = 85 x 0.447 = 38 m/s

initial time (ti) = 0 s