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2 years ago

I NEED HELP ON PHYSICAL SCIENCE; WORK

Physics

2 answers:

Volgvan2 years ago

7 0

Answer: 300 joules

Explanation:

w=f*d

1.5*200=300 joules

Hope this helps plz mark brianliest!

arlik [135]2 years ago

3 0

The answer is 300J hope this helps

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I NEED HELP PLEASE, THANKS! :)

makkiz [27]

Answer:

They move outwards.

They don't intersect each other at any point.

They show the electric field.

Explanation:

8 0

2 years ago

Read 2 more answers

Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0

2 years ago

An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the

garri49 [273]

Answer:

a) u = 30.29 m/s

b) t = 2.09 s

Explanation:

given,

velocity = 45 m/s

angle (θ) = 50°

horizontal velocity = 45 cos 50°

time taken to reach 150 m.

times = $\dfrac{150}{45 cos 50^0}$

t = 5.19 s

a) height of arrow

$s = u t +\dfrac{1}{2}gt^2$

$s = v sin \theta \times t+\dfrac{1}{2}gt^2$

$s = 45 sin 50^0 \times 5.19 -\dfrac{1}{2}\times 9.81\times 5.19^2$

s = 46.78 m

v² - u² = 2 g s

u² = 2 × 9.81 × 46.78

u = 30.29 m/s

b) time taken by the apple = $\dfrac{u}{g}=\dfrac{30.29}{9.81}$

= 3.09 s

time after which it has to be thrown = 5.19-3.09 = 2.1 s

5 0

2 years ago

Queremos un cilindro de simple efecto que utilice en su funcionamiento un volumen de aire a presión atmosférica de 13,122 litros

zhuklara [117]

Answer:

1) El diámetro es de aproximadamente 913,987 cm.

2) La fuerza del cilindro es 5576850 kgf

Explanation:

1) Los parámetros dados son;

El volumen del aire = 13,122 litros = 13122000 cm³

La presión de trabajo = 8.5 kgf / cm²

La longitud del cilindro = 20 cm.

Por lo tanto, tenemos;

El área de la base del cilindro = π · r² = 13122000 cm³ / (20 cm) = 656100 cm²

r = √ (656100 / π) ≈ 456,994 cm

El diámetro = 2 × r ≈ 2 × 456.994 ≈ 913.987 cm

El diámetro ≈ 913,987 cm

2) La fuerza del cilindro = El área de la base del cilindro × La presión de trabajo

∴ La fuerza del cilindro = 656100 cm² × 8.5 kgf / cm² = 5576850 kgf

La fuerza del cilindro = 5576850 kgf

3 0

2 years ago

a 100g ice cube at 0 degrees celsius is placed in 650 grams of water at 25 degrees celsius. When the mixture reaches equillibriu

Artyom0805 [142]

Answer:

The latent heat of fusion of water is 334.88 Joules per gram of water.

Explanation:

Let the latent heat of ice be 'x' J/g

1) Thus heat absorbed by 100 gram of ice to get converted into water equals

$Q_1=100\times x$

2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals

$Q_2=100\times 4.186\times 11=4604.6Joules$

Thus total energy needed equals $Q_1+Q_2=100x+4604.6$

3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is

$Q_3=650\times 4.186\times (25-11)\\\\Q_{3}=38092.6Joules$

Now by conservation of energy we have

$Q_1+Q_2=Q_3\\\\100x+4604.6=38092.6\\\\\therefore x=\frac{38092.6-4604.6}{100}=334.88J/g$

6 0

2 years ago