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Afina-wow [57]
3 years ago
7

On a straight road with the +x axis chosen to point in the direction of motion, you drive for 5 hours at a constant 20 miles per

hour, then in a few seconds, you speed up to 60 miles per hour and drive at this speed for 1 hour.
What was the x component of average velocity for the 6-hour period, using the fundamental definition of average velocity, which is the displacement divided by the time interval?
Physics
1 answer:
Leviafan [203]3 years ago
8 0

Answer:

v = 26.7 mph

Explanation:

During the first 5 hours, at a constant speed of 20 mph, we find the total displacement to be as follows:

Δx₁ = v₁*t₁ = 20 mph*5 h = 100 mi

Assuming we can neglect the displacement during the speeding up from 20 to 60 mph, we can find the the total displacement at 60 mph as follows:

Δx₂ = v₂*t₂ = 60 mph*1 h = 60 mi

So, the total displacement during all the trip wil be:

Δx = Δx₁ + Δx₂ = 100 mi + 60 mi = 160 mi

So we can find the the average velocity during the 6-hour period, applying the definition of average velocity, as follows:

v = Δx / Δt = 160 mi / 6 h = 26.7 mph

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Artist 52 [7]

A) -2.0 m

Look at the ray diagram attached in the picture, where:

p identifies the location of the object

q identifies the location of the image

F identifies the focus of the mirror

Each tick represents 1 m

We have

p = 6.0 m is the distance of the object from the mirror

f = -3.0 m is the focal length

From the ray diagram, we see that q has a distance of 2.0 m from the mirror, and it's on the other side of the mirror compared to the object, so

q = -2.0 m

This can also be verified by using the mirror equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-3.0 m}-\frac{1}{6.0 m}=-\frac{3}{6.0 cm}\\q = \frac{-6.0 cm}{3}=-2.0 cm

B) Upright and virtual

As we see from the picture, the image is upright, since it has same orientation as the object.

Also, we notice that the image is on the other side of the mirror, compared to the object. For a mirror,

- An image is said to be real if it is on the same side of the object

- An image is said to be virtual if it is on the opposite side of the mirror

Therefore, this means that the image is virtual.

8 0
3 years ago
Recycled plastics cannot be used to produce new food containers because ___
Lena [83]

they cannot be sterilized

4 0
3 years ago
The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca
d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}

the elastic potencial energy is given by:

U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J

The work is defined as:

W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s

3 0
3 years ago
Shay reacts solid zinc and aqueous copper sulfate to form aqueous zinc sulfate and solid copper. If he reacts 10.1 grams of zinc
zhenek [66]

Answer:

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

Explanation:

As we know that zinc reacts with copper sulfate

so the reaction is given as

Zn + CuSO_4 --> ZnSO_4 + Cu

so here we have

Zn = 10.1 g

CuSO_4 = 18.6 g

ZnSO_4 = 20 g

Cu = 8.7 g

Now total mass of reactant is given as

M_1 = 10.1 + 18.6 = 28.7 g

Mass of the product is given as

M_2 = 20 + 8.7 = 28.7 g

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

7 0
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How does the electric force vary with distance
wlad13 [49]
The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
3 0
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