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oksian1 [2.3K]
3 years ago
15

How is your guys day going?​

Physics
2 answers:
vichka [17]3 years ago
7 0

Answer:

Awesome and u made it even better when you gave me these free points

Explanation:

faltersainse [42]3 years ago
4 0
Good i’m tired how about you
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A 3.5 kg object gains 76 J of potential energy as it is lifted vertically. Find the new height of the object?
Eduardwww [97]

Answer:

1.72 m

Explanation:

Potential energy = mgh, where m is mass, g is acceleration due to gravity (9.8), and h is height

76 = (3.5)(9.8)h

76=44.1h

h=1.72335600907 ≈1.72 m

8 0
3 years ago
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A man pushes a 10-kg block 10 m, along a rough, horizontal
abruzzese [7]

Answer:

hope you can understand

8 0
3 years ago
A model rocket has an acceleration of 12 m/s2. A net force of 18 N is acting on the rocket. What is the mass of the rocket? its
larisa86 [58]

Answer:

C.) 1.5 kg

Explanation:

Start with the equation:

F_n_e_t=ma

Plug in what you know, and solve:

18=m(12)\\m=1.5kg

Find matching soluation:

C.) 1.5 kg

5 0
2 years ago
A box with mass m is dragged across a level floor with coefficient of kinetic friction μk by a rope that is pulled upward at an
Komok [63]

Answer:

a) F =  μk  mg Cosθ

b) F = 279.78 N

Explanation:

a) F = μk R

Based on the description in the question, the horizontal reaction is:

R = mg Cosθ

The force required to move the box with constant speed in terms of m, μk, θ, and g is :

F =  μk  mg Cosθ

b) If m = 90 kg

g = 9.8 m/s²

μk=0.35

θ = 25⁰

Force required to slide the 90-kg patient across a floor at constant speed by pulling on him at an angle of 25∘ above the horizontal will be:

F =  μk  mg Cosθ

F = 0.35 * 90 * 9.8 * cos25

F = 279.78 N

5 0
4 years ago
A cord is wrapped around the rim of a solid uniform wheel 0.300 m in radius and of mass 8.00 kg. A steady horizontal pull of 34.
disa [49]

Answer:

A. α = 94.4 rad/s

B. a = 28.32 m/s

C. N = 34N

D. α = 94.4 rad/s

    a = 28.32 m/s

     N =  44.4  N

Explanation:

part A:

using:

∑T = Iα

where T is the torque, I is the moment of inertia and α is the angular momentum.

firt we will find the moment of inertia I as:

I = \frac{1}{2}MR^2

Where M is the mass and R is the radius of the wheel, then:

I = \frac{1}{2}(8 kg)(0.3 m)^2

I = 0.36 kg*m^2

Replacing on the initial equation and solving for α, we get::

∑T = Iα

Fr = Iα

34 N =  0.36α

α = 94.4 rad/s

part B

we need to use this equation :

a = αr

where a is the aceleration of the cord that has already been pulled off and r is the radius of the wheel, so replacing values, we get:

a = (94.4)(0.3 m)

a = 28.32 m/s

part C

Using the laws of newton, we know that:

N = T

where N is the force that the axle exerts on the wheel part and T is the tension of the cord

so:

N = 34N

part D

The anly answer that change is the answer of the part D, so, aplying laws of newton, it would be:

-Mg + N +T = 0

Then, solving for N, we get:

N = -T+Mg

N = -34 + (8 kg)(9.8)

N =  44.4  N

6 0
3 years ago
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