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zmey [24]
2 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]2 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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A 5⁢kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant.
Umnica [9.8K]

Answer:

The gravitational force exerted on the object is 75 N (answer D)

Explanation:

Hi there!

The gravitational force is calculated as follows:

F = m · g

Where:

F = force of gravity.

m = mass of the object.

g = acceleration due to gravity (unknown).

For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · a · t²

Where:

y = position at time t.

y0 = initial position.

v0 = initial velocity.

t = time.

g = acceleration due to gravity.

Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.

Then, after 2 seconds, the height of the object will be -30 m:

y = y0 + v0 · t + 1/2 · g · t²

-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²

-30 m = 1/2 · g · 4 s²

-30 m = 2 s ² · g

-30 m/2 s² = g

g = -15 m/s²

Then, the magnitude of the gravitational force will be:

F = m · g

F = 5 kg · 15 m/s²

F = 75 N

The gravitational force exerted on the object is 75 N (answer D)

Have a nice day!

8 0
2 years ago
Use the periodic table and your knowledge of isotopes to complete these statements.
klasskru [66]
<span><span>Use the periodic table and your knowledge of isotopes to complete these statements.

When polonium-210 emits an alpha particle, the child isotope has an atomic mass of </span><span> ⇒ 206</span>.</span>

<span><span>I-131 undergoes beta-minus decay. The chemical symbol for the new element is </span><span> ⇒ Xe</span>.</span>

<span><span>Fluorine-18 undergoes beta-plus decay. The child isotope has an atomic mass of </span><span> ⇒ 18</span>.</span>

7 0
2 years ago
Read 2 more answers
A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle
liq [111]

Answer:

0.0031792338 rad/s

Explanation:

\theta = Angle of elevation

y = Height of balloon

Using trigonometry

tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta

Differentiating with respect to t we get

\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft

Now y = 2500 ft

cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974

\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s

The angle is changing at 0.0031792338 rad/s

6 0
2 years ago
3.Cuanto Calor pierden 514 ml de agua si su temperatura desciende de 12°C a 11°C. Expresa el resultado en calorias.
Scrat [10]

Answer:

514 cal

Explanation:

In order to calculate the lost heat by the amount of water you first take into account the following formula:

Q=mc(T_2-T_1)         (1)

Q: heat lost by the amount of water = ?

m: mass of the water

c: specific heat of water = 1cal/g°C

T2: final temperature of water = 11°C

T1: initial temperature = 12°C

The amount of water is calculated by using the information about the density of water (1g/ml):

m=\rho V=(1g/ml)(514ml)=514g

Then, you replace the values of all parameters in the equation (1):

Q=(514g)(1cal/g\°C)(11\°C-12\°C)=-514cal

The amount of water losses a heat of 514 cal

3 0
2 years ago
Two blocks, with masses m and 3m, are attached to the ends of a string with negligible mass that passes over a pulley, as shown
olasank [31]

Answer:

a)  v = √ g x , b)  W = 2 m g d , c)    a = ½ g

Explanation:

a) For this exercise we use Newton's second law, suppose that the block of mass m moves up

            T-W₁ = m a

            W₃ - T = M a

            w₃ - w₁ = (m + M) a

            a = (3m - m) / (m + 3m) g

            a = 2/4 g

            a = ½ g

the speed of the blocks is

          v² = v₀² + 2 ½ g x

          v = √ g x

b) Work is a scalar, therefore an additive quantity

light block s

           W₁ = -W d = - mg d

3m heavy block

             

            W₂ = W d = 3m g d

the total work is

             W = W₁ + W₂

             W = 2 m g d

c) in the center of mass all external forces are applied, they relate it is

                      a = ½ g

8 0
1 year ago
Read 2 more answers
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