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Bob is pulling a 30kg filing cabinet with a force of 200n , but the filing cabinet refuses to move. the coefficient of static fr

puteri [66]

The cabinet is being pulled with 200N and is being rested by a force equal to 200N. That is why it is not being moved.

<span>Although the force of static friction can equal Fk=µs*F=m*g*µs=(30kg)*(9.8m/s^2)*(0.80)=235 N. It is not resisting the 200N force with 235N. Imagine if you pushed something with 200N and it pushed you back with 235N, especially a cabinet. You would think that the cabinet was alive.</span>

8 0

3 years ago

3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between

Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

$v^2 -u^2 = 2ad$ (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

$F=ma=-\mu mg$

where

m = 1000 kg is the mass of the car

$\mu = 0.80$ is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

$a=-\mu g$

And we can substitute it into eq.(1) to find d:

$v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m$

7 0

3 years ago

A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

Zielflug [23.3K]

Answer: $F_{v} =\mu_{k} mg$

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to start crate moving is $F_{s} =\mu_{s} mg$ .

but once crate starts moving the force of friction is reduced $F_{v} =\mu_{k} mg$ .

Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie $F_{v} =\mu_{k} mg$ .

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.

Magnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

4 0

3 years ago

IP The x and y components of a vector r⃗ are rx = 16 m and ry = -8.5 m , respectively

fiasKO [112]

as it is given that

$r_x = 16 m$

$r_y = -8.5 m$

now we will have

$\vec r = 16 \hat i - 8.5 \hat j$

now the magnitude of this vector is given as

$|r| = \sqrt{16^2 + 8.5^2}$

$|r| = 18 m$

now to find the direction we can use

$tan\theta = \frac{r_y}{r_x}$

$tan\theta = \frac{-8.5}{16}$

$\theta = tan^{-1}(-0.53)$

$\theta = - 28^0$

4 0

3 years ago

Object A is placed on top of object B. Object A is the same temperature as object B. How will heat flow between object A and obj

mash [69]

"No heat will flow between object A and object B" is the one among the following choices given in the question that describes how heat will <span>flow between object A and object B. The correct option among all the options that are given in the question is the second option or option "B". I hope it helps you.</span>

6 0

3 years ago

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