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3 years ago

6

An object is thrown with a force of 30 Newtons and ends up with an acceleration of 3 m/s ^ 2 due to that throw. What is the mass

of that ball?

1 answer:

jonny [76]3 years ago

4 0

Answer:

F=30N

a= 3m/s^2

m=?

F=ma

30=m(3)

30/3=m

m=10kg

The mass of the ball is 10kg

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ruslelena [56]

<span>The answer to this problem is magnesium. I hoped I helped someone with this</span>

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3 years ago

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An airplane is heading due south at a speed of 560 km/h . If a wind begins blowing from the southwest at a speed of 80.0 km/h (a

polet [3.4K]

Answer:

the magnitude of Vpg = 493.711 km/h

Explanation:

given data

speed Vpg = 560 km/h

speed Vwg = 80 km/h

solution

we get here magnitude of the plane velocity w.r.t. ground is

we know that the Vpg = Vpw + Vwg .....................1

writing the component of the velocity that is

Vpw = (0 km/h î - 560 km/h j )

Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)

adding these

Vpg = (0+80 cos 45 km/h ) î + ( -560 + 80 sin 45 km/h j)i

Vpg = (42.025 ) î (-491.92 km/h)j

now we take magnitude

the magnitude of Vpg = $\sqrt{(42.025^2+(-491.92)^2)} km/h$

the magnitude of Vpg = 493.711 km/h

5 0

2 years ago

A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex

Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

$\text{Efficiency of the heat engine} = \eta = 26\% = 0.26$

$\text{Energy taken in by the heat engine during each cycle} = Q_h$

$\text{Energy exhausted by the heat engine in each cycle} = Q_c = 8.55*10^3 J$

$\eta = 1 - \frac{Q_{c}}{Q_{h}}$

$0.26 = 1 - \frac{8.55\ast 10^{3}}{Q_{h}}$

$\frac{8.55* 10^{3}}{Q_{h}} = 0.74$

$Q_h = \frac{8.55*10^3}{0.74}$

$Q_h = 11.554*10^3J$

PART A)

Work done by the heat engine in each cycle = W

$W = Q_h-Q_c$

$W = 11.554*10^3J-8.55*10^3J$

$W = 3004J$

According to the value given we have that,

$P = 4.0kW$

$P = 4000W$

Power is defined as the variation of energy as a function of time therefore,

$P = \frac{W}{t}$

$4000W = \frac{3004J}{t}$

$t = \frac{3004}{4000}$

$t = 0.75s$

Therefore the interval for each cycle is 0.75s

5 0

3 years ago

What is the mass of the object being measured?

ioda

Answer:

m = 375 [gram]

Explanation:

A triple Beam balance is an instrument very easy to use, since we only have to perform the arithmetic sum of each of the weights that are recorded in each beam

m = 300 + 70 + 5 = 375 [gram]

For a better understanding, the following image is attached, with values on each beam, which should be read.

The largest mass is in the indicator of 100 [gram], the second mass is in the indicator of 20 [gram] and the third is in the indicator of 5.8 [gram]. Thus the arithmetic sum corresponds to:

M= 100 + 20 + 5.8 = 125.8 [gram]

Note: it is important that when the instrument is in balance, the opposite end of the beam should indicate a position of zeros.

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3 years ago

You are asked to design a spring that will give a 1020 kg satellite a speed of 2.25 m/s relative to an orbiting space shuttle. Y

julsineya [31]

Answer:

(a) 2.45×10⁵ N/m

(b) 0.204 m

Explanation:

Here we have that to have a velocity of 2.25 m/s then the relationship between the elastic potential energy of the spring and the kinetic energy of the rocket must be

Elastic potential energy of the spring = Kinetic energy of the rocket

$\frac{1}{2} kx^2 = \frac{1}{2} mv^2$

Where:

k = Force constant of the spring

x = Extension of the spring

m = Mass of the rocket

v = Velocity of the rocket

Therefore,

$\frac{1}{2} kx^2 = \frac{1}{2} \times 1020 \times 2.25^2$

or

$kx^2 = 1020 \times 2.25^2 = 10,226.25\\So \ that \ the \ force \ on \ the \ satellite\ kx = \frac{10226.25}{x}$

(b) Since the maximum acceleration is given as 5.00×g we have

Maximum acceleration = 5.00 × 9.81 = 49.05 m/s²

Hence the force on the rocket is then;

Force = m×a = 1020 × 49.05 = ‭50,031 N

$kx = \frac{10226.25}{x} = 50031 \ N$

Therefore,

$x = \frac{10226.25}{ 50031} = 0.204 \ m$

(a) From which

$k = \frac{10226.25}{x^2} = \frac{50031}{x} = \frac{50031}{0.204} = 244,772.13 \ N/m$ or

Force constant of the spring, k = 2.45×10⁵ N/m.

6 0

3 years ago