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Charra [1.4K]
3 years ago
14

A robot standing on a cliff shoots a ball upwards with an initial speed of 30 m/s. What is the height of the cliff given that th

e ball reaches the bottom of the cliff 8 s after the shoot? (Take g = 10 m/s^2 and the height of the robot is negligible.)
A 25 m
B 45 m
C 80 m
D 145 m​

Physics
1 answer:
valina [46]3 years ago
5 0

Answer:

C 80 m

Explanation:

Given:

v₀ = 30 m/s

a = -10 m/s²

t = 8 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²

Δy = -80 m

The ball lands 80 m below where it started.  So the height of the cliff is 80 m.

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Groundwater is the water found underground in the cracks and spaces in soil, sand and rock. It is stored in and moves slowly through geologic formations of soil, sand and rocks called aquifers.


Hope this helps.


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4 0
2 years ago
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A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?
OLga [1]

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

7 0
3 years ago
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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
True or false the hotter the star the higher is absolute magnitude?
liubo4ka [24]

Answer:

true

Explanation:

6 0
3 years ago
A .100 kg bullet is flying at 200 m/s. How much energy is the bullet storing due to its motion?
morpeh [17]

KE = 2000 J

Explanation:

KE = (1/2)mv^2

= (1/2)(0.100 kg)(200 m/s)^2

= 2000 J

4 0
2 years ago
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