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zmey [24]
3 years ago
15

Sally is pushing a shopping cart with a force of 20 N. Because the wheels are stuck, the friction caused by the ground is exerti

ng a force of 8 N in the opposite direction. Calculate the net force on the shopping cart including an arrow indicating what direction the shopping cart is going
Physics
1 answer:
juin [17]3 years ago
0 0

The net force on the shopping cart is 12 N to the right.

This is a question related to Newton's Laws. It can be solved by using a Free body diagram, which shows all the forces acting on the object. There are 4 forces acting on the object:

  1. Gravity (g) pointing down.
  2. Normal (N) pointing up.
  3. Push force by Sally (Fp) pointing right.
  4. Friction force (Ff) pointing left.

Forces (1) and (2) cancel each other because they have the same magnitude, and the overall force is given by the addition of forces (3) and (4)

F=Fp+Ff=20N+ (-8N)=12N

The positive value indicates that the shopping cart is moving to the right.

Have a nice day!

#LearnwithBrainly

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Answer: D

Experiment 1 has a confounding variable related to the mass of the rockets. Any variation in mass may cause a discrepancy in the distance traveled.

This is the answer to the question because:

  • Both experiments do have a confounding variable.
  • Experiment 1 doesn't have to stay constant.
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The answer has to be the option D. Hope this helps you!

3 0
3 years ago
An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be
lord [1]

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity  ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0
3 years ago
Which of the following is true about Viscosity of liquids:
White raven [17]
Viscosity of liquids is essentially the 'thickness' of the liquid. For instance, honey and water have different viscosities. Honey has a higher one and therefore, liquids with high viscosity do not flow as well as liquids with low viscosity (water).
4 0
3 years ago
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jeka57 [31]

Answer:

Stationary Front, warm front, cold front, Occluded Front.

Explanation:

Stationary Front. When the surface position of a front does not change (when two air masses are unable to push against each other; a draw), a stationary front is formed.

cold front is the leading edge of a cooler mass of air at ground level that replaces a warmer mass of air and lies within a pronounced surface trough of low pressure. It often forms behind an extratropical cyclone (to the west in the Northern Hemisphere, to the east in the Southern), at the leading edge of its cold air advection pattern—known as the cyclone's dry "conveyor belt" flow. Temperature differences across the boundary can exceed 30 °C (86 °F) from one side to the other. When enough moisture is present, rain can occur along the boundary. If there is significant instability along the boundary, a narrow line of thunderstorms can form along the frontal zone. If instability is weak, a broad shield of rain can move in behind the front, and evaporative cooling of the rain can increase the temperature difference across the front. Cold fronts are stronger in the fall and spring transition seasons and weakest during the summer.

A warm front is a density discontinuity located at the leading edge of a homogeneous warm air mass, and is typically located on the equator-facing edge of an isotherm gradient. Warm fronts lie within broader troughs of low pressure than cold fronts, and move more slowly than the cold fronts which usually follow because cold air is denser and less easy to remove from the Earth's surface. This also forces temperature differences across warm fronts to be broader in scale. Clouds ahead of the warm front are mostly stratiform, and rainfall gradually increases as the front approaches. Fog can also occur preceding a warm frontal passage. Clearing and warming is usually rapid after frontal passage. If the warm air mass is unstable, thunderstorms may be embedded among the stratiform clouds ahead of the front, and after frontal passage thundershowers may continue. On weather maps, the surface location of a warm front is marked with a red line of semicircles pointing in the direction of travel.

In meteorology, an occluded front is a weather front formed during the process of cyclogenesis. The classical view of an occluded front is that they are formed when a cold front overtakes a warm front, such that the warm air is separated (occluded) from the cyclone center at the surface. The point where the warm front becomes the occluded front is called the triple point; a new area of low-pressure that develops at this point is called a triple-point low. A more modern view of the formation process suggests that occluded fronts form directly during the wrap-up of the baroclinic zone during cyclogenesis, and then lengthen due to flow deformation and rotation around the cyclone.

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julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

F_k = F_{W,E}

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The extension of the spring due to the weight of the object on Moon is a value of x_2, then

kx_2 = mg_m

Recall that gravity on the moon is a sixth of Earth's gravity.

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kx_2 = \frac{1}{6} kx_1

x_2 = \frac{1}{6} x_1

We have that the displacement at the earth was x_1 = 0.3m, then

x_2 = \frac{1}{6} 0.3

x_2 = 0.05m

Therefore the displacement of the mass on the spring on Moon is 0.05m

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