Multiple Choice Questions Please do all relevant calculations in the sheet given

Question

Multiple Choice Questions Please do all relevant calculations in the sheet given an Answer ALL problems using the bubble sheet given support your answer with i Can an objects velocity change direction when its acceleration is constant? example down, but it can never tum A) No, it is not possible ecause it is always speeding up slowing B) No, this is not possible because it is always speeding up or always around and backs up is an example D) Yes, this possible, and rock thrown up from rest, speeds up, slows to a stop, is possible, and a car that 2. The density of seawater was measured to be 1.07 g/cm This density in SI units is a) 1.07 x 10 kg/m (/1.07 x 10 kg/m c) 1.07 x 10 kg 1.07 x 10 kg. e) 1.07 x 10 kg/m ors A, B, C have the following y components: and 3. Three x component +6 -3 +2 component -3 +4 +5 The angle that the resultant makes with the positive direction oftheraxis is A) 1.2 B) 36° C) 500 D) 40 E) 70e 4. If the speed of that ofparticie B, the distance particle B travels in a given interval of time as compared with particle A A) Four times as great B) Twice as great D) Half as great E) one-fourth as great 5. A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0s. It then has a uniform acceleration of 10 cm/s for another 5.0 s. The particle moves in the same direction along a straight line. The average speed over the whole time interval is A) 20 cm/s B) 35 cm/s D) 100 cm/s E) 12 cm/s

Explanation / Answer

way too many questions.. here are the first 8

1. D)

2) 1.07 g/cm^3 = 1070 kg/m^3
so c)

3)
net y = 6
net x = 5

arctan(y/x) = arctan(6/1)= 50
so c)

4) d= v t so double v doubles d

so B)

5)

so distance in first section
v = v0 + at
0 = .3 + a 5
a = -.3/5
v^2 = v0^2 + 2 a x
0 = .3^2 -2*.3/5*x
x=0.75 m

second interval
x = 1/2 at^2 = 0.5*.1*5^2= 1.25
xnet = 2

so v average = 2/10 = 0.2 = 20 cm/s
so a)

6) x = 1/2 a t^2
so double t 4 times x

so d)

7) c)

8) b)

9) ax = 7.1/0.25

ay = 4.8/0.25

a = sqrt( (7.1/0.25)^2 +(4.8/0.25)^2)= 34

and noth positive so c)

10)
y direction
y = y0 + v0y t + 1/2 a t^2

0 = 600-250*sin(37 degrees)*t - 0.5*9.81*t^2
t=3.57 s

x = 250*cos(37 degrees)*3.57= 713 m
so b)

15)

a = 3.5/0.7

sum forces
t - mg = ma

T = 40*(9.81 + 3.5/0.7)= 592 N

so a)

16)

t cos theta - u N = m a

frction = T cos theta - ma

so b)

17) so b = 3.2e-2/16

terminal when bv = mg

(3.2E-2/16)*v = 12.0E-3*9.81
v= 59 m/s

so b)

18)

Fnet = 2 N jhat
mnet = 15

so a =Fnet/mnet = 2/15 jhat
so C)

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