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Helen [10]
2 years ago
13

The sum of all the forces acting on an object is zero.

Physics
2 answers:
sergey [27]2 years ago
8 0

Answer:

Equilibrium -The sum of all the forces acting on the object is zero, which is the first condition of it and it has two forms. Law of Inertia -The body will remain at rest or move at constant velocity unless acted. upon by an external net force or unbalanced force.

Explanation:

:) tell me if it's wrong please!

marysya [2.9K]2 years ago
3 0

Answer:

Newton's first law states that when the vector sum of all forces acting on an object (the net force) is zero, the object is in equilibrium. If the object is initially at rest, it remains at rest. If it is initially in motion, it continues to move with constant velocity.

Explanation:

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A soccer player runs with a speed of 4.6m/s how long does it take him to run 60m
Misha Larkins [42]

To solve this problem divide 60 by 4.6

The answer to this problem is 13 seconds.

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3 years ago
Impulse equals?<br> A) momentum x velocity<br> B) momentum x time<br> C) mass x velocity
mina [271]

Answer:

B

Explanation:

The impulse experienced by an object is the force•time.

7 0
3 years ago
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Calculate the acceleration of a student riding his bicycle in a straight line that speeds up from 4 m/s to 6 m/s in 5 seconds.
solong [7]
Acceleration is equal to velocity final minus velocity initial divided by time. 6m/s minus 4m/s divided by 5 seconds is 0.4m/s^2.

7 0
3 years ago
Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0° with the horizo
Aleks [24]

Answer: Given mass of man = 85kg

Distance crate was pushed = 4m

Force exerted on crate = 500N

Angle inclined = 20°

Let Total force exerted Ft = (mg * sin(20)) + 500

Ft =( 85*9.81* sin20°) + 500

Ft = 758N

Work done Wd = Ft * distance

Wd = 785 * 4 = 3.14 * 10^3J

3 0
3 years ago
An object is thrown with an initial velocity v0 forming an angle θ with an inclined plane, which a In turn it forms an α-angle α
xxMikexx [17]
Refer to the figure shown below, which is based on the given figure.

d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.

Part A
From the geometry, obtain
d = X cos(α)                     (1a)
h = X sin(α)                      (1b)

The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α)               (2a)
u = v₀ cos(θ - α)             (2b)

If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0             (3a)
ut = d                                (3b)

Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain
0.5(9.8)( \frac{d}{u})^{2} -v_{0} sin(\theta -  \alpha ) \frac{d}{u} - h = 0
4.9[ \frac{X cos \alpha }{v_{0} cos(\theta -  \alpha }  ]^{2} - v_{0} sin(\theta -  \alpha ) [ \frac{X cos \alpha }{v_{0} cos(\theta -  \alpha } ] - X sin \alpha  = 0
Hence obtain
aX^{2}-bX=0 \\ where \\ a=4.9[ \frac{cos \alpha }{v_{0} cos(\theta -  \alpha )}]^{2} \\  b = cos \alpha \,  tan(\theta -  \alpha ) + sin \alpha
The non-triial solution for X is
X= \frac{b}{a}

Answer:
X= \frac{sin \alpha  + cos \alpha  \, tan(\theta -  \alpha )}{4.9 [ \frac{cos \alpha }{v_{0} \, cos(\theta -  \alpha )}  ]^{2}}

Part B
v₀ = 20 m/s
θ = 53°
α = 36°

sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423

X = 0.8351/(4.9*0.0423²) = 101.46 m

Answer:  X = 101.5 m

7 0
3 years ago
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