PLEASE HELP!! THIS IS DUE RN!!

: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

e-lub [12.9K]

Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

$a_s = \dfrac{F}{m}$

$a_s = \dfrac{8.2}{12}$

$a_s = 0.68\ m/s^2$

$F = m a_m$

$a_m = \dfrac{F}{m}$

$a_m = \dfrac{8.2}{70}$

$a_m = 0.12\ m/s^2$

$x_1+x_2 = 25$

$\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25$

$(a_c+a_m)t^2=50$

$(0.12+0.68)t^2=50$

$t = \sqrt{\dfrac{50}{0.8}}$

t = 7.90 s

$x_1 = \dfrac{1}{2}a_ct^2$

$x_1 = 0.5\times 0.12 \times 7.90^2$

$x_1 = 3.74m$

5 0

3 years ago

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

Alex

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = $m_1=6.50 kg- 1.55 kg=4.95 kg$

Velocity achieved by squid = $v_1=2.60 m/s$

Momentum gained by squid = $P=m_1v_1$

Mass of water = $m_2=1.55 kg$

Velocity by which water was released by squid = $v_2$

Momentum gained by water but in opposite direction = $P'=m_2v_2$

P = P'

$m_1v_1=m_2v_2$

$v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s$

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E = $\frac{1}{2}m_1v_1^{2}$

Kinetic energy of water = K.E' = $\frac{1}{2}m_2v_2^{2}$

Total kinetic energy created by this maneuver:

$K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}$

$=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules$

4 0

3 years ago

A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant

mixas84 [53]

Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

3 0

3 years ago

Fusion and fission are both processes that involve which fundamental force? to ask? A. gravitational B. weak nuclear C. electrom

marin [14]

Fusion & Fission are both processes that involve D. Strong Nuclear Fundamental forces
"D". is the answer

5 0

3 years ago

An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1

Ksivusya [100]

<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

We have gravitational force

$F=\frac{GMm}{r^2}$

Where G = 6.67 x 10⁻¹¹ N m²/kg²

M = Mass of body 1

M = Mass of body 2

r = Distance between them

Here we have

M = Mass of Sun = 1.99×10³⁰ kg

m = Mass of asteroid = 4.00×10¹⁶ kg

F = 3.14×10¹³ N

Substituting

$F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m$

The asteroid is 4.11 x 10¹¹ m far from Sun

3 0

3 years ago