1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
nadya68 [22]
1 year ago
6

Driving safely at night requires seeing well not only in low light conditions, but also being able to see low contrast

Physics
1 answer:
Monica [59]1 year ago
8 0

Answer:

True

Explanation:

Driving safely at night requires seeing well not only under low light, but also requires drivers to see low-contrast objects.

You might be interested in
According to a general theory of crime, individuals who engage in crime are also likely to engage in other risky behaviors, for
babymother [125]

Answer:

construct validity

Explanation:

According to my research on studies conducted by various psychologists, I can say that based on the information provided within the question it seems that the researcher is using construct validity. This is the act of measuring or validating what the extent to which a claim is true. In this situation since the claim states that people involved in crimes tend to do speed and drugs, then the researcher compares the criminals responses to those areas to test the validity of the claim.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

6 0
3 years ago
a desktop computer and monitor together draw about 2 A of current they plug into a wall outlet that is 120 V what is the Resista
Delvig [45]

Answer:

60 \Omega

Explanation:

the relation between current, voltage and resistance in an electrical circuit is given by Ohm's law:

V=IR

where V is the voltage, I is the current and R is the resistance. In this problem, the current is I=2 A, the voltage is V=120 V, therefore we can arrange the previous equation and find the resistance:

R=\frac{V}{I}=\frac{120 V}{2 A}=60 \Omega

7 0
2 years ago
A conveyor belt is used to move sand from one place to another in a factory. The conveyor is tilted at an angle of 18° above the
brilliants [131]

Answer:

x = 2.044 m

Explanation:

given data

initial vertical component of velocity = Vy = 2sin18

initial horizontal component of velocity = Vx = 2cos18

distance from the ground yo = 5m

ground distance y = 0

from equation of motion

y = yo+ V_y t +\frac{1}{2}gt^2

0 = 5 + 2sin18+ \frac{1}{2}*9.8t^2

solving for t

t = 1.075 sec

for horizontal motion

x = V_x t

x = 2cos18*1.075

x = 2.044 m

8 0
3 years ago
Any wanna talk to me
Sliva [168]

Answer:hello

Explanation:

6 0
2 years ago
A glass lying on table doesn't possess friction.why?
Ratling [72]
Becsud it's not moving
7 0
3 years ago
Other questions:
  • While at a party, you pull up a sound intensity level app on your phone (everyone does stuff like that, right?), and it reads 83
    8·1 answer
  • If the mass of a material is 104 grams and the volume of the material is 16 cm3, what would the density of the material be?
    6·1 answer
  • which of the following would not be taken into consideration when describing the quality of a sound? a. the number of the overto
    10·2 answers
  • How does the kinetic energy of an object relate to its mass and velocity?
    14·1 answer
  • In what part of the scientific method do you record data?
    9·1 answer
  • Can someone please help me answer my question I’m really confused and I have a test tomorrow please and thank you god bless
    13·1 answer
  • Help plsss i need #1
    14·1 answer
  • One relationship between organisms is that of predator-prey. Which of the following is the best description of a predator?
    12·2 answers
  • Help!! Urgent!! What property determines the sequence of layers in the earth?
    6·1 answer
  • Two 5000-kg passenger cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track. They co
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!