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Roman55 [17]
3 years ago
11

Scott is walking 11 m/s due south along a road when he spots Karla, who is walking toward him at 1.5 m/s. On her way, Karla sees

Homer and Juanita walking together at 1.3 m/s. For which people can a
velocity be determined?
Scott and Karla
B
Homer and Juanita
C
Karla and Juanita
D
Homer and Scott
Physics
1 answer:
emmasim [6.3K]3 years ago
8 0

Answer: A. Scott and Karla

Explanation: Velocity includes speed AND direction.

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Question is in the picture below
Sati [7]

may vary depending on the organization.

4 0
3 years ago
Parker (73.2 kg) is being dragged down the hall with an applied force of 123 N. If the frictional force is 27.4 N, what is the c
levacccp [35]

Answer:

The coefficient of friction in the hall is 0.038

Explanation:

Given;

mass of the Parker, m = 73.2 kg

applied force on the parker, F = 123 N

frictional force, Fs = 27.4 N

the coefficient of friction in the hall = ?

frictional force is given by;

Fs = μN

Where;

μ is the coefficient of friction

N is normal reaction = mg

Fs = μmg

μ = Fs / mg

μ = (27.4) / (73.2 x 9.8)

μ = 0.038

Therefore, the coefficient of friction in the hall is 0.038

6 0
3 years ago
Identify which type of source is being described.
frez [133]

Answer:

Primary, secondary

Explanation:

3 0
3 years ago
Read 2 more answers
Hello!
34kurt
Iodine-131 has a half life of 8 days, so half of it is gone every 8 days.
10 grams of iodine-131 is left for 24 days.
At 8 days: 10/2=5 grams left
At 16 days: 5/2=2.5 grams left
At 24 days: 2.5/2=1.25 grams left.
**
Your mistake is that you stopped at 16 days.
8 0
3 years ago
An 8.5 kg crate is pulled 5.1 m up a 30 degree incline by a rope angled 17 degrees above the incline. The tension in the rope is
Ulleksa [173]

Answer:

1. a W_t=746.63 J

  b E_p=212.415 J

  c W_n=183.96J

2. T_e=99.71J

Explanation:

a). The work done by the tension is:

W_t=T*dt

dt=\frac{5.1}{cos(17)}

W_t=140N*\frac{5.1}{cos(17)}

W_t=746.63 J

b). The work done potential of gravity

E_p=m*g*h

h=5.1*sin(17)

E_p=8.5kh*9.8*5.1*sin(30)

E_p=212.415 J

c). The work done by the normal force

W_n=N*d_n

d_n=5.1*sin(30)=2.55

W_n=8.5kg*9.8*cos(30)*2.55

W_n=183.96J

2. The increase in thermal energy is:

T_e=F*d

F_k=u_k*m*g=0.271*8.5kg*9.8*cos(30)

F_k=19.5N

T_e=19.55*5.1m

T_e=99.71J

4 0
3 years ago
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