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OLga [1]
3 years ago
5

Lincoln weighs 400 newtons. What’s his mass rounded to the nearest kilogram? Assume that acceleration due to gravity is 9.8 N/kg

.
Physics
2 answers:
Bogdan [553]3 years ago
8 0

Answer:

m = 40.8 [kg]

Explanation:

We know that weight is defined as the product of mass by gravitational acceleration in this way we have the following equation.

W = m*g

where:

W = weight [N] (units of Newtons)

m = mass [kg]

g = gravity acceleration = 9.8 [N/kg]

Therefore:

m=W/g\\m=400/9.8\\m=40.8[kg]

Sonbull [250]3 years ago
8 0

Answer: 41!

Explanation:

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Dumbo the elephant weights 13,000 pounds. What force would the Earths surface apply to Dumbo? 2.2 pounds = 1 kg
snow_lady [41]

Answer:

57908 N

Explanation:

Let's first convert Dumbo's mass into kg using the given relationship: 2.2 pounds =1 kg.

Then, 13000 lbs = 13000/2.2 kg = 5909 kg

Now, let's find the force of gravity on Dumbo at the surface of the earth, which would be in magnitude equal to the normal force that the Earth's surface applies on Dumbo.

F = m * a = 5909 kg * 9.8 m/s^2 = 57908 N

8 0
2 years ago
SIEVERT (SV) IS THE PRODUCT OF ABSORBED DOSE AND RADIATION WEIGHTING FACTOR<br> T True<br> F False
Alla [95]

Answer:

False

Explanation:

Sievert is the unit of dose equivalent

4 0
2 years ago
3. The smallest bird is the Cuban bee hummingbird, which has a mass of only
Firlakuza [10]

Answer:

2.6 m

Explanation:

The work done by the bird is given by

W=Fd

where

F is the force exerted

d is the distance covered

In this problem, we know:

W=8.8\cdot 10^{-4} J is the work

F=3.4\cdot 10^{-4} N is the force

Solving the equation for d, we find the distance covered by the bird:

d=\frac{W}{F}=\frac{8.8\cdot 10^{-4}}{3.4\cdot 10^{-4}}=2.6 m

8 0
3 years ago
The force P is applied to the 45-kg block when it is at rest. Determine the magnitude and direction of the friction force exerte
cluponka [151]

Answer:

Check attachment for free body diagram of the question.

I used the free body diagram and the angles given in the diagram missing in the question above, but I used the data given in the above question.

Explanation:

Let frictional force be Fr acting down the plane

Let analyze the structure before inserting values

Using Newton's second law along the y-axis

ΣFy = Fnet = m•ay

Since the body is not moving in the y-direction, then ay = 0

N+PSinβ — WCosθ = 0

N+PSin20—441.45Cos15 = 0

N+PSin20—426.41 = 0

N = 426.41 — PSin20 , equation 1

The maximum Frictional force to be overcome is given as

Fr(max) = μsN

Fr(max) = 0.25(426.41 — PSin20)

Fr(max)= 106.6 —0.25•PSin20

Fr(max) = 106.6 — 0.08551P, equation 2

This is the maximum force that must be overcome before the body starts to move

Using Newton's law of motion in the x direction

Note, we took the upward direction up the plane as the direction of motion since the force want to move the block upward

Fnetx = ΣFx

Fnetx = P•Cosβ —W•Sinθ — Fr

Fnetx = P•Cos20—441.45•Sin15—Fr

Fnetx = 0.9397P — 114.256 — Fr

Equation 3

When Fnetx is positive, then, the body is moving up the plane, if Fnetx is negative, then, the maximum frictional force has not yet being overcome and the object is still i.e. not moving

a. When P = 0

From equation 2

Fr(max) = 106.6 — 0.08551P

Fr(max) = 106.6 — 0.08551(0)

Fr(max)= 106.6 N

So, 106.6N is the maximum force to be overcome

So, here the only force acting on the body is the weight and it acting down the plane, trying to move the body downward.

Wx = WSinθ

Wx = 441.45× Sin15

Wx = 114.256 N.

Since the force trying to move the body downward is greater than the maximum static frictional force, then the body is not in equilibrium, it is moving downward.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — PSin20 , equation 1

N = 426.41 N, since P=0

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 426.41

Fr = 93.81 N.

b. Now, when P = 190N

From equation 2

Fr(max) = 106.6 — 0.08551(190)

Fr(max) = 106.6 —16.2469

Fr(max)= 90.353 N

So, 90.353 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 190Cos20 — 441.45Sin15

Fnetx = 64.29N

So the force moving the body up the incline plane is 64.29N

Fnetx < Fr(max)

Then, the frictional force has not being overcome yet.

Then, the body is in equilibrium.

Then, applying equation 3.

Fnetx = 0.9397P — 114.256 — Fr

Fnetx = 0, since the body is not moving

0 = 0.9397(190) —114.246 — Fr

Fr = 64.297 N

Fr ≈ 64.3N

c. When, P = 268N

From equation 2

Fr(max) = 106.6 — 0.08551(268)

Fr(max) = 106.6 —16.2469

Fr(max)= 83.68 N

So, 83.68 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 268Cos20 — 441.45Sin15

Fnetx = 137.58 N

So the force moving the body up the incline plane is 137.58 N

Fnetx > Fr(max)

Then, the frictional force has being overcome.

Then, the body is not equilibrium.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — 268Sin20 , equation 1

N = 334.75 N, since P=268N

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 334.75

Fr = 73.64 N

d. The required force to initiate motion is the force when the block want to overcome maximum frictional force.

So, Fnetx = Fr(max)

Px — Wx = Fr(max)

From equation 1

Fr(max) = 106.6 — 0.08551P,

P•Cosβ-W•Sinθ = 106.6 — 0.08551P

P•Cos20 — 441.45•Sin15 = 106.6 — 0.08551P

P•Cos20—114.256=106.6 - 0.08551P

PCos20+0.08551P =106.6 + 114.256

1.025P=220.856

P = 220.856/1.025

P = 215.43 N

3 0
3 years ago
Consider a satellite in a circular orbit around the Earth. If it were at an altitude equal to twice the radius of the Earth, 2RE
Elenna [48]

Answer:

v=\sqrt{\frac{gR_E}{2}}

Explanation:

Satellites experiment a force given by Newton's Gravitation Law:

F=\frac{GMm}{r^2}

where M is Earth's mass, m the satellite's mass, r the distance between their gravitational centers and G the gravitational constant.

We also know from Newton's 2nd Law that <em>F=ma, </em>so putting both together we will have:

ma=\frac{GMm}{r^2}

a=\frac{GM}{r^2}

If we are on the surface of the Earth, the acceleration would be g and r=R_E (Earth's radius):

g=\frac{GM}{R_E^2}

Which we will write as:

gR_E^2=GM

If we are on orbit the acceleration is centripetal (a=\frac{v^2}{r}), so we have:

\frac{v^2}{r}=a=\frac{GM}{r^2}=\frac{gR_E^2}{r^2}

v^2=\frac{gR_E^2}{r}

v=\sqrt{\frac{gR_E^2}{r}}

And if this orbit has a radius r=2R_E we have:

v=\sqrt{\frac{gR_E^2}{2R_E}}=\sqrt{\frac{gR_E}{2}}

3 0
3 years ago
Read 2 more answers
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