A rocket sled accelerates at a rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body.

Answers

Answer 1

Answer:

Explanation:

It is given that,

Mass of the passenger, m = 75 kg

Acceleration of the rocket, $a=49\ m/s^2$

(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :

F = m a

$F=75\ kg* 49\ m/s^2$

F = 3675 N

Ratio, $R=(F)/(W)$

$R=(3675)/(75* 9.8)=5$

So, the ratio between the horizontal force and the weight is 5 : 1.

(b) The magnitude of total force the seat exerts against his body is F' i.e.

$F'=√(F^2+W^2)$

$F'=√((3675)^2+(75* 9.8)^2)$

F' = 3747.7 N

The direction of force is calculated as :

$\theta=tan^(-1)((W)/(F))$

$\theta=tan^(-1)((1)/(5))$

$\theta=11.3^(\circ)$

Hence, this is the required solution.

Answer 2

Answer:

Final answer:

The horizontal component of the force the seat exerts against the passenger's body is 3675 N. The ratio of this force to the passenger's weight is 5. The total force the seat exerts has a magnitude of 3793 N.

Explanation:

(a) To calculate the horizontal component of the force the seat exerts against the passenger's body, we can use Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the passenger is 75.0 kg and the acceleration of the rocket sled is 49.0 m/s2. So the force exerted by the seat is:

Force = mass * acceleration

Force = 75.0 kg * 49.0 m/s2

Force = 3675 N

Now let's compare this force to the passenger's weight. The weight of an object is given by the formula:

Weight = mass * gravitational acceleration

Weight = 75.0 kg * 9.8 m/s2

Weight = 735 N

To find the ratio, we divide the force exerted by the seat by the weight of the passenger:

Ratio = Force / Weight

Ratio = 3675 N / 735 N

Ratio = 5

(b) The total force the seat exerts against the passenger's body has both a horizontal and vertical component. The direction of the total force is the same as the direction of the acceleration of the rocket sled. The magnitude of the total force can be found using the Pythagorean theorem:

Total Force = √(horizontal component2 + vertical component2)

Total Force = √(36752 + 7352)

Total Force = 3793 N

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Answers

Answer:

Explanation:

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A plastic ball in a liquid is acted upon by its weight and by a buoyant force. The weight of the ball is 4 N. The buoyant force has a magnitude of 5 N and acts vertically upward. When the ball is released from rest, what is it's acceleration and direction? [2 pts] for a Free Body Diagram correctly labeled.

Answers

Answer:

The acceleration is 2.448 meters per square second and is vertically upward.

Explanation:

The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:

$\Sigma F = F - m\cdot g = m\cdot a$ (1)

Where:

$F$ - Buoyant force, measured in newtons.

$m$ - Mass of the plastic ball, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration, measured in meters per square second.

If we know that $F = 5\,N$ , $m = 0.408\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then the net acceleration of the plastic ball is:

$a = (F)/(m) - g$

$a= 2.448\,(m)/(s^(2))$

The acceleration is 2.448 meters per square second and is vertically upward.

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine provides anet thrust of 0.795 N perpendicular tothe tethering wire.(a) Find the torque the net thrust producesabout the center of the circle.
N·m

(b) Find the angular acceleration of the airplane when it is inlevel flight.
rad/s2

(c) Find the linear acceleration of the airplane tangent to itsflight path.
m/s2

Answers

(a) 24.6 Nm

The torque produced by the net thrust about the center of the circle is given by:

$\tau = F r$

where

F is the magnitude of the thrust

r is the radius of the wire

Here we have

F = 0.795 N

r = 30.9 m

Therefore, the torque produced is

$\tau = (0.795 N)(30.9 m)=24.6 N m$

(b) $0.035 rad/s^2$

The equivalent of Newton's second law for a rotational motion is

$\tau = I \alpha$

where

$\tau$ is the torque

I is the moment of inertia

$\alpha$ is the angular acceleration

If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

$I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2$

And so we can solve the previous equation to find the angular acceleration:

$\alpha = (\tau)/(I)=(24.6 Nm)/(707.5 kg m^2)=0.035 rad/s^2$

(c) $1.08 m/s^2$

The linear acceleration (tangential acceleration) in a rotational motion is given by

$a=\alpha r$

where in this problem we have

$\alpha = 0.035 rad/s^2$ is the angular acceleration

r = 30.9 m is the radius

Substituting the values, we find

$a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2$

7) Straws work on the principle of the outside atmospheric pressure pushing the fluid (for example water) up the straw after you have lowered the pressure at the top of the straw (in your mouth). Assuming you could create a perfect vacuum in your mouth, what is the longest vertical straw you could drink water from?

Answers

Answer:

The longest straw will be 10.328 meters long.

Explanation:

The water will rise up to a height pressure due to which will balance the atmospheric pressure.

We know

$P_(atm)=101325N/m^(2)$

Pressure due to water column of height 'h'

$P_(water)=1000* 9.81* h$

Equating both the values we get the value of height 'h' as

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Answers

Answer:

Gravity

Explanation:

Gravity is constantly pulling objects downward. Without it, everything would float out into space.

I hope this answer helps :)

Answer:

The answer for the given question above would be option C. GRAVITATIONAL FORCE. Based on the given scenario above of a leaf that falls to the ground when Tonya let it go, the force that pulled the leaf to the ground is the gravitational force. This kind of force is a force that attracts any object with mass.

Hope this helps!!!

wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength

Answers

Complete question:

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:

a) the midpoint between the two rings?

b) the center of the left ring?

Answer:

a) the electric field strength at the midpoint between the two rings is 0

b) the electric field strength at the center of the left ring is 2712.44 N/C

Explanation:

Given;

distance between the two rings, d = 25 cm = 0.25 m

diameter of each ring, d = 10 cm = 0.1 m

radius of each ring, r = $(0.1)/(2) = 0.05 \ m$

the charge on each ring, q = 20 nC

Electric field strength for a ring with radius r and distance x from the center of the ring is given as;

$E = (kxQ)/((x^2 +r^2)^(3/2))$

The electric field strength at the midpoint;

the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m

$E = (kxQ)/((x^2 +r^2)^(3/2)) \n\nE = (8.99*10^(9)*0.125*20*10^(-9))/((0.125^2 + 0.05^2)^(3/2)) \n\nE = 9210.5 \ N/C$

$E_(left) = 9210.5 \ N/C$

The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;

$E_(right) = -9210.5 \ N/C$

The electric field strength at the midpoint;

$E_(mid) = E_(left) + E_(right)\n\nE_(mid) = 9210.5 \ N/C - 9210.5 \ N/C\n\nE_(mid) = 0$

(b)

The distance from the right ring to center of the left ring, x = 0.25 m.

$E = (KxQ)/((x^2 +r^2)^(3/2)) \n\nE = (8.99*10^(9) *0.25*20*10^(-9))/((0.25^2 + 0.05^2)^(3/2)) \n\nE = 2712.44 \ N/C$

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