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2 years ago

a 5.4 kg bag of groceries is in equilibrium on an incline of angle θ = 15°. find the magnitude of the normal force on the bag

Physics

1 answer:

Sever21 [200]2 years ago

5 0

Answer: 51.2 N

Explanation:

The magnitude of the normal force acting on the bag counterbalances the component of the weight which is perpendicular to the incline, equal to:

$W_p = mg cos \theta$

where

m = 5.4 kg is the mass of the bag

g = 9.81 m/s^2 is the gravitational acceleration

$\theta=15^{\circ}$ is the angle of the incline

Substituting the numbers into the formula, we get

$W_p = (5.4 kg)(9.81 m/s^2 )(cos 15^{\circ}=51.2 N$

And so, the normal force is also equal to 51.2 N.

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A 0.110 kg cube of ice (frozen water) is floating in glycerine. The glycerine is in a tall cylinder that has inside radius 3.70

Sonbull [250]

Answer:

the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

Explanation:

The change in volume of glycerin when the ice cube is placed on the surface of the glycerin can be represented as:

$V = \frac{m}{ \rho}$

Given that ;

the mass of the ice cube (m) = 0.11 kg = 0.110 × 10³ g

density of the glycerine ( $\rho$ ) = 1.260 kg/L = 1.260 g/cm³

Then:

V = $\frac{0.110*10^3 \ g}{1.260 \ g/cm^3}$

V = $0.0873*10^3 \ cm^3 (\frac{1L}{10^3 cm^3})$

V = 0.0873 L

Now;Initially the volume of the glycerin before the ice cube starts to melt is:

$V_1 = V_i + V\\\\V_1 = V_i+ 0.0873 \ L$

However; the volume of the water produced by the 0.11 kg ice cube = $0.11*10^3 \ cm^3$

The expression for change in the volume of glycerin after the ice cube starts to melt is as follows:

$V_2 = V_i + V"$

replacing $V"$ with $0.11*10^3 \ cm^3$ ; we have:

$V_2 = V_i (0.11*10^3 \ cm^3 )(\frac{1 \ L }{10^3 \ cm^3})$

$V_2 = V_i + 0.11 \ L$

The overall total change in the volume of the glycerin is illustrated as:

$V_f = V_2 - V_1$

Now; from the foregoing ; lets replace the respective value of $V_2$ and $V_1$ in the above equation ; we have;

$V_f = (V_i + 0.11 \ L) - (V_i + \ 00873 \ L)\\ \\V_f = 0.11 L - 0.0873 \ L\\\\V_f = 0.0227 \ L$

The formula usually known to be the volume of a cylinder is :

$V = \pi r ^2 h$

For the question ; we will have:

$V_f = \pi r ^2 h$

making h the subject of the formula ; we have:

$h = \frac{V_f}{\pi r^2}$

replacing 0.0227 L for $V_f$ and the given value of radius which is = 3.70 cm; we have:

$h = \frac{0.0227 \ L ( \frac{10^3 \ cm^3}{1\ L})}{\pi * (3.70 cm)^2}$

$h = \frac{22.7 \ cm^3}{\pi * (3.70 cm)^2}\\\\h = 0.528 \ \ cm$

Thus ; the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

8 0

2 years ago

The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig

julia-pushkina [17]

Answer: They are all true

a. The tension in the rope is everywhere the same.

b. The magnitudes of the forces exerted on the two objects by the rope are the same.

c. The forces exerted on the two objects by the rope must be in opposite directions.

d. The forces exerted on the two objects by the rope must be in the direction of the rope.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)

- Cutiepatutie ☺❀❤

5 0

2 years ago

You use a slingshot to launch a potato horizontally from the edge of a cliff with speed v0. The acceleration due to gravity is g

Ray Of Light [21]

Answer:

$\displaystyle t=\frac{2v_o}{g}$

Explanation:

<u>Horizontal Launch</u>

When an object is launched horizontally at a speed vo, it describes a curved called parabola as the speed in the x-direction does not change and the speed in the y-direction increases with time because the gravity makes it return to the ground.

The vertical distance the object (potato) travels downwards is:

$\displaystyle y=\frac{gt^2}{2}$

The horizontal distance is

$x=v_ot$

We need to find the time when both distances are equal, thus

$\displaystyle \frac{gt^2}{2}=v_ot$

Simplifying by t

$\displaystyle \frac{gt}{2}=v_o$

Solving for t

$\displaystyle \boxed{t=\frac{2v_o}{g}}$

8 0

2 years ago

g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the

Svetlanka [38]

Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

$R = \frac{u^{2}Sin2\theta }{g}$

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

$R = \frac{u^{2}Sin60 }{9.8}$

R = 0.088u^2

If the angle of projection is 45 degree.

$R = \frac{u^{2}Sin90 }{g}$

R = u^2 / g

5 0

2 years ago

If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =

Butoxors [25]

(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.

3 0

2 years ago