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kkurt [141]
1 year ago
14

Init. A

Physics
1 answer:
Gekata [30.6K]1 year ago
7 0

Answer:

v = 1/3 m / s = 0.333 m / s

in the direction of the truck

Explanation:

The average speed is defined by the variation of the position between the time spent

           v = Δx / Δt

since the position is a vector we must add using vectors, we will assume that the displacement to the right is positive, the total displacement is

           Δx = 20 - 15 +20

           Δx = 25 m

therefore we calculate

         v = 25/75

         v = 1/3 m / s = 0.333 m / s

in the direction of the truck

You might be interested in
Two students walk in the same direction along a straight path, at a constant speed one at 0.90 m/s and the other at 1.90 m/s. a.
creativ13 [48]

Answer: a) 456.66 s ; b) 564.3 m

Explanation: The time spend to cover any distance a constant velocity is given by:

v= distance/time so t=distance/v

The slower student time is: t=780m/0.9 m/s= 866.66 s

For the faster students t=780 m/1,9 m/s= 410.52 s

Therefore the time difference is 866.66-410.52= 456.14 s

In order to calculate the distance that faster student should  walk

to arrive 5,5 m before that slower student, we consider the follow expressions:

distance =vslower*time1

distance= vfaster*time 2

The time difference is 5.5 m that is equal to 330 s

replacing in the above expression we have

time 1= 627 s

time2 = 297 s

The distance traveled is 564,3 m

8 0
2 years ago
Sayid made a chart listing data of two colliding objects. A 5-column table titled Collision: Two Objects Stick Together with 2 r
Alborosie

Answer:

6 m/s is the missing final velocity

Explanation:

From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).

Object X had a mass of 300 kg, while object Y had a mass of 100 kg.

Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.

We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.

In numbers, and calling P_{xi} the initial momentum of object X and P_{yi} the initial momentum of object Y, we can derive the total initial momentum of the system: P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}

Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

Final momentum of the system: M * v_f=400kg * v_f

We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):

2400 \frac{kg*m}{s} =400kg*v_f\\\frac{2400}{400} \frac{m}{s} =v_f\\v_f=6 \frac{m}{s}

7 0
2 years ago
Read 2 more answers
What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t
Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

\Delta P =  \frac{8 \mu L Q}{\pi r^4}

where:

\Delta P is the pressure difference between the two ends

\mu is viscosity of the fluid

L is the length of the pipe

Q=Av is the volumetric flow rate, with A=\pi r^2 being the section of the tube and v the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

\mu=0.001 Pa/s is the dynamic water viscosity at 20^{\circ}

L=4.0 cm=0.04 m

Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s

and r=1 mm=0.001 m

Using these data in the formula, we get:

\Delta P = 3200 Pa

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0
1 year ago
Calculate the buoyant force in air on a kilogram of titanium (whose density is about 4.5 grams per cubic centimeter). compare wi
aleksklad [387]
1) The buoyant force acting on an object immersed in a fluid is:
B=d_f V_d g
where d_f is the density of the fluid, V_d is the volume of displaced fluid, and g=9.81~m/s^2 is the gravitational acceleration.

2) We must calculate the volume of displaced fluid. Since the titanium object is completely immersed in the fluid (air), this volume corresponds to the volume of 1 Kg of titanium, whose density is d=4.5~g/cm^3 = 4.5\cdot10^3~Kg/m^3. Using the relationship between density, volume and mass, we find
V_d= \frac{m}{d}= \frac{1~Kg}{4.5\cdot10^3Kg/m^3}=2.22\cdot10^{-4}~m^3

3) Now we can recall the formula written at step 1) and calculate the buoyant force. The air density is d_f = 1~Kg/m^3, so we have
B=d_f V_d g=1~Kg/m^3 \cdot 2.22\cdot10^{-4}~m^3 \cdot 9.81~m/s^2=2.22\cdot10^{-3}~N

4) The weight of 1 Kg of titanium is instead:
W=mg=1~Kg \cdot 9.81~m/s^2=9.81~N
So, the buoyant force is negligible compared to the weight.
7 0
1 year ago
It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you
Juli2301 [7.4K]

Answer:

0.0984

Explanation:

From the first diagram attached below; a free flow diagram shows the interpretation of this question which will be used  to solve this question.

From the diagram, the horizontal component of the force is:

F_X = F_{cos \ \theta}

Replacing 42°  for θ and 87.0° for F

F_X =87.0 \ N \ *cos \ 42 ^\circ

F_X =64.65 \ N

On the other hand, the vertical component  is ;

F_Y = Fsin \ \theta

Replacing 42°  for θ and 87.0° for F

F_Y =87.0 \ N \ *sin \ 42 ^\circ

F_Y =58.21  \ N

However, resolving the vector, let A be the be the component of the mutually perpendicular directions.

The magnitude of the two components is shown in the second attached diagram below and is now be written as A cos θ and A sin θ

The expression for the frictional force is expressed as follows:

f = \mu \ N

Where;

\mu is said to be the coefficient of the friction

N = the  normal force

Similarly the normal reaction (N) = mg - F sin θ

Replacing F_Y \ for \ F_{sin \  \theta}. The normal reaction can now be:

N = mg \ - \ F_Y

By balancing the forces, the horizontal component of the force equals to frictional force.

The horizontal component of the force is given as follows:

F_X = \mu \ ( mg - \ F_Y)

Making \mu the subject of the formular in the above equation; we have the following:

\mu \ = \ \frac{F_X}{mg - F_Y}

Replacing the following values: i.e

F_X \ = \ 64.65 \  N

m = 73 Kh

g  = 9.8 m/s²

F_Y = \ 58.21 N

Then:

\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}

\mu = 0.0984

Thus, the coefficient of friction is = 0.0984

5 0
1 year ago
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