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vodka [1.7K]
2 years ago
8

The speed of an object undergoing constant acceleration increased from 8.0 meters per second to 16.0 meters per second in 10. Se

conds . How far does the object travel during the 10. Seconds
Physics
1 answer:
saw5 [17]2 years ago
6 0

v₀ = initial speed of the object = 8 meter/second

v = final speed of the object = 16 meter/second

t = time taken to increase the speed = 10 seconds

d = distance traveled by the object in the given time duration = ?

using the kinematics equation

d = (v + v₀) t/2

inserting the above values in the above equation

d = (16 + 8) (10)/2

d = 120 meter


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Why does frost bite occur in mountain climbers ?​
san4es73 [151]

Answer:

The higher the altitude the colder

Explanation:

Mountains are very high up obviously higher then the ground. It is colder so it makes it easier for mountain climbers to get frost bite because of the high altitude.

Hope this helps:)

7 0
2 years ago
Read 2 more answers
High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 43.7 m/s just before it strikes a 45
Helga [31]

Answer:

41.27m/s

Explanation:

According to law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the velocity after impact

Given

m1 = 0.2kg

u1 = 43.7m/s

m2 = 45.9g = 0.0459kg

u2 = 30.7m/s

Required

Velocity after impact v

Substitute the given parameters into the formula

0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v

8.74+1.409 = 0.2459v

10.149 = 0.2459v

v = 10.149/0.2459

v = 41.27m/s

Hence the speed of the golf ball immediately after impact is 41.27m/s

8 0
2 years ago
A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity
Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2  = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches  = 0.0508 m

n = 1750 rpm

H_{nom}=2hp=1491.4W

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m

V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s

F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N

(F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N

T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm

F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N

F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N

b)

H_a=1491*1.25=1863.75W

n_f_s=\frac{H_a}{H_{nom}K_S }=1

dip = \frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm

7 0
2 years ago
A 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Afterwards, the 355 kg car moves right at 5.80
Sidana [21]

Explanation:

It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

202\ kg\times 8.5\ m/s+355\ kg\times 0=m_1v_1+355\ kg\times 5.8\ m/s

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

7 0
2 years ago
The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through alumi
Alex777 [14]

Answer:

(A) The correct answer is option (B) three halves that of the old unit.

(B) The answer is three fourth that of old unit

Explanation:

from the relation;

(A) Fromthe expression;

K = Qd/AΔT

Anew = 3/2 A(old)

(B)

K¹ = 2K(old), so we get

A(old) = A(old)/2

Combining with part A, we have;

Anew = 3/2 *A(old)/2

          = 3/4A(old)

The answer is three fourth that of old unit

8 0
2 years ago
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