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Nonamiya [84]
3 years ago
9

Andrew is pushing this block to the right across on the floor. Which force in the free body diagram is pointing in the wrong dir

ection?
Question 11 options:

the applied force


friction


the normal force


the weight

Physics
2 answers:
soldier1979 [14.2K]3 years ago
8 0

Answer:

the normal force

Explanation:

The free-body diagram represents all the forces acting on an object. In this example, there are four forces acting on the box: an applied force, the friction (which always act opposite to the applied force), the weight of the box (which is always downward), and the normal force.

The normal force is the reaction force exerted by the surface on which the box is moving on the box, and this reaction force is always opposite to the force exerted by the box on the surface. Since the latter is downward, it means that the normal force must be upward, so in the diagram it is wrong.

Deffense [45]3 years ago
6 0

Normal force since it's perpendicular to the contact force and should be going up.

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A water bath in a physical chemistry lab is 1.55 m long, 0.710 m wide, and 0.570 m deep (high). If it is filled to within 3.55 i
Lesechka [4]

Answer:

528 liter.

Explanation:

Volume of the tank(cuboid) = l*b*h

But volume of the water = l*b*h

Where

l= length of the tank

b = width of the tank

h = the length from the bottom of the tank,

3.55 in to m,

0.09017m

Length of the water in the tank = 0.570 - 0.09017

= 0.47983 m.

Volume = 0.47983*0.710*1.55

= 0.528 m3.

1 m3 = 1000 liter.

0.528 m3 = 0.528*1000

= 528 liter

7 0
3 years ago
an ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component
Nata [24]

The y-component of the acceleration is 0.28 m/s^2

Explanation:

The y-component of the ice skater acceleration can be calculated with the equation

a_y = \frac{v_y-u_y}{t}

where

v_y is the y-component of the final velocity

u_y is the y-component of the initial velocity

t is the time elapsed

Here we have:

  • Initial velocity is u=2.25 m/s at \theta_1=50.0^{\circ}, so its y-component is u_y = u sin \theta_1 = (2.25)(sin 50.0^{\circ})=1.72 m/s
  • Final velocity is v=4.65 m/s at \theta_2=120.0^{\circ}, so its y-component is v_y = v sin \theta_2 = (4.65)(sin 120.0^{\circ})=4.03 m/s

The time elapsed is

t = 8.33 s

Therefore, the y-component of the acceleration is

a_y = \frac{4.03-1.72}{8.33}=0.28 m/s^2

Learn more about acceleration:

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3 years ago
A square-based shipping crate is being designed that must contain a volume of 16 ft3 . The material that is used for the base an
Vlada [557]

Answer:

Explanation:

Given

volume V=16 ft^3

Suppose base is square with side L

height of crate is h

Volume V=L^2\times h

16=L^2\times h

Cost of top and bottom area c_1=3L^2

Cost of Side area c_2=4Lh\times 2=8Lh=8L\times \frac{16}{L^2}=\frac{128}{L}

Total Cost C=c_1+c_2

Total Cost C=3L^2+\frac{128}{L}

Differentiate C w.r.t Length

\frac{dC}{dL}=6L-\frac{128}{L^2}

L^3=\frac{128}{6}

L=2.75 ft

h=\frac{16}{2.75^2}=11.46 ft

Dimensions are L\times L\times h=2.75\times 2.75\times 11.46    

6 0
3 years ago
A boy drops a 0.10 kg stone down a 150 m well and listens for the echo. The air temperature is 20°C. How long after the stone is
S_A_V [24]
Use formula for Echo which is Velocity=2(Distance)/time so 343=2(150)/T 343T=300..T=300/343=0.9 seconds
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3 years ago
A target behavior is
crimeas [40]

Answer:

Target Behavior would be the behavior known to be changed, it can be defined by function or topography

3 0
3 years ago
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