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gladu [14]
2 years ago
12

PLEASEE HELP I !!!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
Ratling [72]2 years ago
5 0
The answer was is B
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What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a
Naya [18.7K]

Answer:

T= 38.38 N

Explanation:

Here

mass of can = m = 3 kg

g= 9.8 m/sec2

angle θ = 40°

From figure we see the vertical and horizontal component of tension force T

If the can is to slip - then horizontal component of tension force should become equal to force of friction.

First we find force of friction

Fs= μ R

where

μ = 0.76

R = weight of can = mg = 3 × 9.8 = 29.4 N

Now horizontal component of tension

Tx= T cos 40 = T× 0.7660  N

==>T× 0.7660 = 29.4

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8 0
3 years ago
H. A truck starts to move from rest. If it gains the acceleration of 3 m/s2 in 5 sec,
SashulF [63]

Answer:

1. The final velocity of the truck is 15 m/s

2. The distance travelled by the truck is 37.5 m

Explanation:

1. Determination of the final velocity

Initial velocity (u) = 0 m/s

Acceleration (a) = 3 m/s²

Time (t) = 5 s

Final velocity (v) =?

The final velocity of the truck can be obtained as follow:

v = u + at

v = 0 + (3 × 5)

v = 0 + 15

v = 15 m/s

Therefore, the final velocity of the truck is 15 m/s

2. Determination of the distance travelled

Initial velocity (u) = 0 m/s

Acceleration (a) = 3 m/s²

Time (t) = 5 s

Distance (s) =?

The distance travelled by the truck can be obtained as follow:

s = ut + ½at²

s = (0 × 5) + (½ × 3 × 5²)

s = 0 + (½ × 3 × 25)

s = 0 + 37.5

s = 37.5 m

Therefore, the distance travelled by the truck is 37.5 m

6 0
2 years ago
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