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Alona [7]
3 years ago
14

Referring to the graph below of data describing a cart pulled across a level surface. When a 4 N force pulls on the cart, the ca

rt accelerates at 5 m/s^2. What is the friction force resisting the pull?

Physics
2 answers:
sashaice [31]3 years ago
5 0

Answer:

0.5 N

Explanation:

(0.6 kg)(9.8 m/s^2) = 5.88 N

(0.6 kg)(5 m/s^2) = 3 N

4 N - 3 N = 1 N

1 N/5.88 N = 0.17

(0.17)(3 N) = 0.51 N which rounds to 0.5 N

LenKa [72]3 years ago
4 0

Answer:gjv

Explanation:

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what happens at Point C is sublimation. the increase in temperature affects the Vapour pressure soon as you can see the curve is increasing with increasing pressure there is increase in temperature that is the sublimation Curve

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A container with rigid walls is filled with 4.0 mol of air with Cv=2.5R Then the temperature is increased from 17 degrees C to 3
galina1969 [7]

Explanation:

Internal energy = heat + work

U = Q + W

Since there's no change in volume (rigid walls), W = 0.

U = Q

U = n Cᵥ ΔT

U = (4.0 mol) (2.5 × 8.314 J/mol/K) (354 C − 17 C)

U = 28,000 J

3 0
3 years ago
A 250 kg cart is at the top of a hill that is 32 m high, what is its potential energy?
atroni [7]

Answer:

<h2>80,000 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 250 × 10 × 32

We have the final answer as

<h3>80,000 J</h3>

Hope this helps you

3 0
3 years ago
How do you calculate the gravitational force on a rubber ball which has a mass of 50g
AfilCa [17]
P (gravitational force) = m (mass) x g
<=> P = 0.05 x 10
<=> P = 0.5N
6 0
3 years ago
A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t
Jobisdone [24]

Answer:

(a) E=0  :   0 cm from the center of the sphere

(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C    :    40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C  :    59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

4 0
3 years ago
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