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4 years ago

A force that pulls objects toward the center of the Earth is called _____.

Physics

1 answer:

Ratling [72]4 years ago

4 0

B.) <span>A force that pulls objects toward the center of the Earth is called "Gravity"

Hope this helps!</span>

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The global winds and moisture belts indicate that large amounts of rainfall occur at

diamong [38]

it should be rising and converging

3 0

3 years ago

What is the magnitude of fs on an object lying on a flat surface without moving, on

Degger [83]

The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

magnitude of force on the object, F = 10 N

angle between the object and the horizontal flat surface = 0⁰

Apply Newton's second law of motion to determine the magnitude of the force on the object.

Due to the position of the object, the magnitude of the force acting on it is calculated as;

$\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N$

Therefore, the magnitude of the force acting on the object is 10 N.

Learn more here: brainly.com/question/19887955

7 0

3 years ago

A 150 g sample of brass at 100 °C is placed in a Styrofoam cup of water containing 120 mL of water at 10 °C. No heat is lost to

fredd [130]

Answer:

≈19.144°C.

Explanation:

all the details are in the attachment.

Note, that c₁, m₁, t₁ are the parameters of the sample of brass; c₂, m₂ and t₂ are the parameters of the sample of water.

P.S. change the provided design according Your requirements.

4 0

2 years ago

The temperature of 1 m^3 of water is decreased by 10°C. If this thermal energy is used to lift the water vertically against grav

Rama09 [41]

Answer:

h = 4271.43 m

Explanation:

given,

Volume of the water = 1 m³

temperature decrease by = 10°C

heat removed from water

Q = m c ΔT

Q = ρ V c ΔT

= 1000 × 1 × 4186 × 10

= 4.186 × 10⁷ J

energy is used to do work to move the water against its weight

Q = force × displacement

4.186 × 10⁷ J = m g × h

4.186 × 10⁷ J = 1000 × 1 × 9.8 × h

h = 4271.43 m

hence, the change in height of is equal to h = 4271.43 m

8 0

4 years ago

7) A kite 50 ft above the ground moves horizontally at a speed of 4 ft/sec. At what rate is the length of the string increasing

d1i1m1o1n [39]

Imagine a right triangle where the legs represent the horizontal and vertical lengths of the string and the hypotenuse represents the length of the string.

Let us assign some values:

x = horizontal length in feet

50 = vertical length in feet

L = length of the string in feet

Because we are modeling these quantities with a right triangle, we can use the Pythagorean theorem to relate them with the following equation:

L² = x² + 50²

We want to find an equation for the change of L over time, so first differentiate both sides with respect to time t then solve for dL/dt:

2L(dL/dt) = 2x(dx/dt)

dL/dt = (x/L)(dx/dt)

First let's solve for x at the moment in time described in the problem using the Pythagorean theorem:

L² = x² + 50²

Given values:

L = 100ft

Plug in and solve for x:

100² = x² + 50²

x = 86.6ft

Now let's find dL/dt. Given values:

x = 86.6ft, L = 100ft, dx/dt = 4ft/sec

Plug in and solve for dL/dt:

dL/dt = (86.6/100)(4)

dL/dt = 3.46ft/sec

3 0

4 years ago