Ask question

Ask question

All categories

2 years ago

10

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take

n as the xy plane. The 4.80-kg puck has a velocity of 1.00î m/s at one instant. Eight seconds later, its velocity is (6.00î + 6.0ĵ) m/s.(a) Assuming the rocket engine exerts a constant horizontal force, find the components of the force.

1 answer:

Ad libitum [116K]2 years ago

7 0

Answer:

$F=(3i+3.6j)\ N$

Explanation:

It is given that,

Mass of the puck, m = 4.8 kg

Initial velocity of the puck, $u=(1i+0j)\ m/s$

After 8 seconds, final velocity of the puck, $v=(6i+6j)\ m/s$

Let the x and y component of force is given by $F_x\ and\ F_y$ .

x component of force is given by :

$F_x=m\times \dfrac{v-u}{t}$

$F_x=4.8\times \dfrac{6-1}{8}$

$F_x=3\ N$

y component of force is given by :

$F_y=m\times \dfrac{v-u}{t}$

$F_y=4.8\times \dfrac{6-0}{8}$

$F_y=3.6\ N$

So, the component of the force is $F=(3i+3.6j)\ N$ . Hence, this is the required solution.

You might be interested in

Which statements describe vectors? Check all that apply. -Vectors have magnitude only. -Vectors have direction only. -Vectors ha

Natali [406]

Answer:

Vectors have both magnitude and direction

Explanation:

Vectors show how strong the force in because the bigger the arrow, the stronger the force. Also, it obviously shows direction because its an arrow.

6 0

2 years ago

Read 2 more answers

A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft b

trasher [3.6K]

Answer:

height of the water rise in tank is 10ft

Explanation:

Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)

$\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2$

$\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}$ -------(1)

substitute $P_a_t_m for P_1, (P_a_t_m +pgh) for P_2$

0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)

$\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}$

$\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}$

Applying bernoulli's equation between tank surface (3) and orifice exit (4)

$\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4$

substitute

$P_a_t_m for P_3, P_a_t_m for P_4$

0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g

$\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}$

At equillibrium Fow rate at point 2 is equal to flow rate at point 4

Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

$\sqrt{64.4(20-h)}$ for V₂

$\sqrt{64.4h}$ for V₄

A₂V₂ = A₃V₃

$A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft$

Therefore ,height of the water rise in tank is 10ft

3 0

2 years ago

Frances drew a diagram to show electromagnetic induction.

kari74 [83]

Answer:

The answer is B) Magnetic field

Explanation:

I chose it and I got it right

8 0

2 years ago

Read 2 more answers

"For a first order instrument with a sensitivity of .4 mV/K and a time" constant of 25 ms, find the instrument’s response as a f

ELEN [110]

Answer:

Explanation:

Given that:

For a first order instrument with a sensitivity of .4 mV/K

constant c = 25 ms = 25 × 10⁻³ s

The initial temperature $T_1$ = 273 K

The final temperature $T_2$ = 473 K

The initial volume = 0.4 mV/K × 273 K = 109.2 V

The final volume = 0.4 mV/K × 473 K = 189.2 V

the instrument’s response as a function of time for a sudden temperature increase can be computed as follows:

Let consider y to be the function of time i.e y(t).

So;

y(t) = 109.2 + (189.2 - 109.2)( 1 - $\mathbf{e^{-t/c}}$ )mV

y(t) = (109.2 + 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

Plot the response y(t) as a function of time.

The plot of y(t) as a function of time can be seen in the diagram attached below.

What are the units for y(t)?

The unit for y(t) is mV.

Find the 90% rise time for y(t90) and the error fraction,

The 90% rise time for y(t90) is as follows:

Initially 90% of 189.2 mV = 0.9 × 189.2 mV

= 170.28 mV

170.28 mV = (109.2 + 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

170.28 mV - 109.2 mV = 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

61.08 mV = 80 ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

0.7635 mV = ( 1 - $\mathbf{e^{t/25\times 10^{-3}}}$ )) mV

t = 1.44 × 25 × 10⁻³ s

t = 0.036 s

t = 36 ms

The error fraction = $\dfrac{189.2-170.28 }{189.2}$

The error fraction = 0.1

The error fraction = 10%

8 0

2 years ago

A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved

MariettaO [177]

Answer:

4.17 m/s

Explanation:

To solve this problem, let's start by analyzing the vertical motion of the pea.

The initial vertical velocity of the pea is

$u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s$

Now we can solve the problem by applying the suvat equation:

$v_y^2-u_y^2=2as$

where

$v_y$ is the vertical velocity when the pea hits the ceiling

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = 1.90 is the distance from the ceiling

Solving for $v_y$ ,

$v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s$

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

$v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s$

Therefore, the speed of the pea when it hits the ceiling is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s$

5 0

2 years ago