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Alexxandr [17]
2 years ago
11

Determine the change in velocity of a car that starts at rest and has a final velocity of 20 m/s north

Physics
1 answer:
Veseljchak [2.6K]2 years ago
4 0
The overall change in velocity is just 20m/s.
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A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦
blagie [28]

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.  

W= F*d *cosα Formula (1)  

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)  

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force,  inclined at  θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the  tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7°  (N)

Calculated of the FN  

We apply the formula (2)  

∑Fy = m*ay ay = 0  

FN +Ty- W = 0  

FN = W-Ty  

FN =  89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*(  89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax  ,  ax= 0 ,because the speed of the cart  is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

6 0
3 years ago
Question 2 (Multiple Choice Worth 3 points)
zmey [24]
Surveys are considered the most reliable way to gather data
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3 years ago
A NASA explorer spacecraft with a mass of 1,000 kg takes off in a positive direction from a stationary asteroid. If the velocity
Jet001 [13]

Before the launch, the momentum of the (spacecraft + asteroid) was zero.  So after the launch, the momentum of the (spacecraft + asteroid) has to be zero.

Momentum = (mass) x (velocity)

Momentum after the launch:

Spacecraft:  (1,000 kg) x (250 m/s) = 250,000 kg-m/s

Asteroid: (mass) x (-25 m/s)

Their sum:  250,000 - 25(mass) .

Their sum must be zero, so  250,000 kg-m/s = (25 m/s) x (mass)

Divide each side by  25 :  10,000 kg-m/s = (1 m/s) x (mass)

Divide each side by (1 m/s) :  10,000 kg = mass


3 0
3 years ago
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H e l p a h o m i e o u t t h a n k s
finlep [7]
Uh it’s D i’m pretty sure
4 0
2 years ago
A pitcher throws a baseball with a mass of 143 g horizontally at a speed of 38.8 m/s (87 mi/h). The hitter's bat is in contact w
SIZIF [17.4K]

Answer:

F = −10093.41 N

Explanation:

Given that,

Mass of a baseball, m = 143 g = 0.143 kg

Initial speed of the baseball, u = +38.8 m/s

The hitter's bat is in contact with the ball for 1.20 ms and then travels straight back to the pitcher's mound at a speed of 45.9 m/s, v = -45.9 m/s

We need to find the average force exerted on the ball by the bat. So, Force is given by :

F=ma

a is acceleration

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.143\times (-45.9-(38.8))}{1.2\times 10^{-3}}\\\\F=-10093.41\ N

So, the average force exerted on the ball by the bat has a magnitude of 10093.41 N.

8 0
3 years ago
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