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aniked [119]
3 years ago
12

Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s

. Sue applies her brakes but can accelerate only at −1.80 m/s2 because the road is wet. Will there be a collision?
Physics
1 answer:
Zielflug [23.3K]3 years ago
7 0

Answer:

there will be collision

Explanation:

v_{s} =  speed of sue = 34 m/s

v_{v} = speed of van = 5.20 m/s

v_{sv} = speed of sue relative to van  = v_{s} - v_{v} = 34 - 5.20 = 28.8 m/s

d_{s} = stopping distance after brakes are applied

D = distance between sue and van = 160 m

v_{f} = final speed of sue = 0 m/s

a = acceleration = - 1.80 m/s²

Using the kinematics equation

v_{f}^{2} = v_{o}^{2} + 2 a d_{s}

0^{2} = 28.8^{2} + 2 (1.80) d_{s}

d_{s} = 230.4 m

Since  d_{s} < D

hence there will be collision

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1. Plot the following graphs:
VLD [36.1K]

Answer:

(a) The distance-time graph for an object with uniform speed is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram

(b) The distance-time graph for an object with non-uniform speed is giving by a curved line sloped graph with varying gradient as shown in the attached diagram

(c) The velocity-time graph for a car with uniform motion is giving by a horizontal line graph at the speed of constant motion with a zero gradient as shown in the attached diagram

(d) The velocity-time graph for a car moving with uniform acceleration is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram

(e) The velocity-time graph for a car moving with non-uniform acceleration is giving by a curved line sloped graph with varying gradient as shown in the attached diagram

(f) According to Newton's first law of motion, an object at rest will remain at rest with no motion unless acted by a force, an therefore, will have no motion with time

Explanation:

8 0
3 years ago
How much charge is on each plate of a 3.00-μF capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is c
Sauron [17]

Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

V = 15 V

(a) q = C x V

where, q be the charge.

q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

6 0
3 years ago
A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H
Aleks04 [339]

Answer:

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

Final KE  =  (1/2) (1430 kg) (11.0 m/s)²  =   86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .

Explanation:

Dont report my answer please

3 0
3 years ago
Please solve this question ​
lesantik [10]

Answer:

88200 Pa

it is because

height =9m

density=1000kg/m(cube)

gravity = 9.8m/s(square)

now,

P=d×g×h

= 1000×9.8×9

=88200pa

8 0
3 years ago
I'm not an idiot so I'm pretty sure c and d are not the answers. I'm confused because don't they both do this. As it picks up sp
krok68 [10]
The answer is A because it's MOVING

ithink
8 0
3 years ago
Read 2 more answers
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