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Inga [223]
3 years ago
12

1. A psychologist exposes several lab animals to an enriched environment (toys to play with) and then compares their ability to

run mazes to that of control animals.
 
a). Hypothesis:
 
b). Independent variable:
 
c). Dependent variable:
Physics
2 answers:
mamaluj [8]3 years ago
5 0
I don’t fully understand the question being asked, but i would say C. Dependent variable since it is exposed to different things.
lilavasa [31]3 years ago
4 0
Hypothesis: Enriched environments will have a positive effect and better performance of the experimental group of animals exposed to toys in comparison to the control group of animals in running in mazes.

Dependent Variable: Control Group

Independent Variable: Exposed/Experimental Group
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A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. At the bottom, 4.0 km farther, his speed is at
Allisa [31]

We are given:

v0 = initial velocity = 18 km/h

d = distance = 4 km

v = final velocity = 75 km/h

a =?

<span>
We can solve this problem by using the formula:</span>

v^2 = v0^2 + 2 a d

 

75^2 = 18^2 + 2 (a) * 4

5625 = 324 + 8a

<span>a = 662.625 km/h^2</span>

6 0
3 years ago
Bodies weighing 1 kilogram and 5 kilograms lie on a smooth horizontal surface. If a traction force of 0.6 N acts on another 5 kg
natima [27]

0.6/5,1,5

so calculate it

not so sure though

6 0
3 years ago
008 (part 3 of 4) 3.0 points
motikmotik

Car A take a time of 2.55hr and car B take a time of 2.14 hr

We know that distance divide by time is speed

here it is given that car A to reach a gas station a distance 189 km from the school traveling at a speed of 74 km/hr​

so speed=distance/time

s=d/t

t=d/s

=189/74

=2.55hr

In case of car B it is given that The distance from the is 199.8km, car b is traveling at a speed of 93 km/hr  

s=d/t

t=d/s

=199.8/93

=2.14hr  

so from the above given data and the formula we solved and found out the time taken by car A is 2.55h and car B is 2.14h

learn more about Speed here brainly.com/question/13943409

#SPJ9

5 0
11 months ago
A ball is shot at an angle of 45 degrees into the air with initial velocity of 46 ft/sec. Assuming no air resistance, how high d
mestny [16]

Answer:

5.02 m

Explanation:

Applying the formula of maximum height of a projectile,

H = U²sin²Ф/2g...................... Equation 1

Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.

Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°

Constant: g = 9.8 m/s²

Substitute these values into equation 1

H = (14.021)²sin²45/(2×9.8)

H = 196.5884×0.5/19.6

H = 5.02 m.

Hence the ball goes 5.02 m high

8 0
2 years ago
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
3 years ago
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