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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 28-kg beginning roller skater is standing in front of a wall. By pushing against the wall, she is propelled backward with a ve
Scorpion4ik [409]

Answer:

33.6 Ns backward.

Explanation:

Impulse: This can be defined as the product of force and time. The S.I unit of impulse is Ns.

From Newton's second law of motion,

Impulse = change in momentum

I = mΔv................................. Equation 1

Where I = impulse, m = mass of the skater, Δv = change in velocity = final velocity - initial velocity.

Given: m = 28 kg, t = 0.8 s, Δv = -1.2-0 = -1.2 m/s (Note: the initial velocity of the skater = 0 m/s)

Substituting into equation 1

I = 28(-1.2)

I = -33.6 Ns

Thus the impulse = 33.6 Ns backward.

3 0
3 years ago
On a trip you travel at a constant speed of 50 kilometers per hour North for 1 hour. Then you turn West and travel at a constant
polet [3.4K]

Answer:27 km per hour West + 17 km per hour North

6 0
2 years ago
(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the
shtirl [24]

Answer:

Direction of ship: 9.45° West of North

Ship's relative speed: 7.87m/s

Explanation:

A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0

Vx=0

Therefore, -VsSin∅+VcCos∅40°

Sin∅ = Vc/Vs × Cos 40°

Sin∅ = 1.5/7 ×Cos40°

Sin∅= 0.164

∅= Sin-¹ (0.164)

∅= 9.45° W of N

B. Ship's relative speed:

Vy= VsCos∅ + Vcsin40°

= 7Cos9.45° + 1.5sin40°

= 7×0.986 + 1.5×0.642

= 7.865

= 7.87m/s

4 0
3 years ago
The sphere that refers to Earth's water is called what?
WARRIOR [948]

Answer:

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Explanation:

6 0
3 years ago
Read 2 more answers
The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the
Lynna [10]

Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

s=ut+0.5at^2

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, a=-10.0 m/s^2

\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is s=44.0 m

The acceleration is same, a=-10.0 m/s^2

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

v^2-u^2=2as

0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.





7 0
3 years ago
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