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Roman55 [17]
3 years ago
7

The chart below summarizes the forces applied to four different objects.

Physics
1 answer:
posledela3 years ago
6 0

Answer:

C. Y

Explanation:

From Newton's second law of motion, we know that:

       Force = mass x acceleration

So;

      acceleration  = \frac{Force }{mass}  

 Therefore, to have the highest acceleration at a constant force, the mass must be low. Acceleration is inversely proportional to mass.

 

Y has the least mass and it will have the highest acceleration

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Physics multiple choice
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10. b

11.d

12. b

13.e

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ANSWER:
C

EXPLANTION:
The earths crust is the toppest layer of Earth, so that elimates option A and B.

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3 years ago
A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards
Mrrafil [7]

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

5 0
3 years ago
Suppose Joe, who can type 40 words per minute, takes 2.5 hours to type his report. To speed things up, his friend Mark offers to
Vika [28.1K]

Explanation:

When Joe works alone, the total number  of words he typed can be given by:

Total words = (40 words per minute) x (60 minutes per hour) x (2.5 hours)

Total words = 6000 words

Now, when Joe and Mark work together, let 'y' be the number of hours for which they both work simultaneously:

Total words = Words Typed by Joe + Words Typed by Mark

6000 = {(40 words per minute) x (60 minutes per hours) x (y hours)} + {(20 words per minute) x (60 minutes per hours) x (y hours)}

6000 = 2400y + 1200y = 3600y

y = 1.67 hours = 1 hour and 40 minutes

Thus, working together simultaneously, Joe and Mark will take 1 hour and 40 minutes to complete the report.

5 0
3 years ago
A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20 m above the ground. Determine the
marusya05 [52]

Answer:

Explanation:

Given

Initial velocity of ball u=10\ m/s

height of window h=20\ m

Using Equation of motion

y=ut+\frac{1}{2}at^2

where u=initial velocity

t=time

a=acceleration

As ball is already is at a height of 20 m so

Y=ut+\frac{1}{2}at^2+20

Y=10\times t+0.5\times (-9.8)t^2+20

Y=-4.9t^2+10t+20

(b)highest point is obtained at v=0

v^2-u^2=2as

where

v=final velocity

u=initial velocity

a=acceleration

s=displacement

(0)-10^2=2\times (-9.8)\times s

s=\frac{100}{19.6}

s=5.102\ m

Highest Point will be s+20=25.102\ m

(c)Time taken when the ball hit the ground i.e. at Y=0

-4.9t^2+10t+20=0

t=3.28\ s

impact velocity v=\sqrt{2\times 9.8\times 25.102}

v=22.181\ m/s

7 0
3 years ago
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