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3 years ago

The roller coaster from problem #1 then tops a second hill at 15.0 m/s, how high is the second hill?

Physics

1 answer:

navik [9.2K]3 years ago

4 0

Answer:

68.5

Explanation:

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Until a train is a safe distance from the station it must travel at 5 m/s. Once the train is on open track it can speed up

kirza4 [7]

Answer:

I believe the answer is b

Explanation:

5 0

4 years ago

Can uh help in in this question step by step

Luda [366]

Initial velocity=u=72km/h

Convert to m/s

$\\ \sf \longmapsto 72\times \dfrac{5}{18}=5(4)=20m/s$

Final velocity=v=0m/s
Time=2s=t

$\\ \sf \longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf \longmapsto Acceleration=\dfrac{0-20}{2}$

$\\ \sf \longmapsto Acceleration=\dfrac{-20}{2}$

$\\ \sf \longmapsto Acceleration=a=-10m/s^2$

Distance be s

Using second equation of kinematics

$\\ \sf \longmapsto s=ut+\dfrac{1}{2}at^2$

$\\ \sf \longmapsto s=20(2)+\dfrac{1}{2}(-10)(2)^2$

$\\ \sf \longmapsto s=40+(-20)$

$\\ \sf \longmapsto s=40-20$

$\\ \sf \longmapsto s=20m$

Now

Mass=m=5000kg

Using newtons second law

$\\ \sf \longmapsto Force=ma$

$\\ \sf \longmapsto Force=5000(-10)$

$\\ \sf \longmapsto Force=-50000N$

Force is in opposite direction so its negative

$\\ \sf \longmapsto Force=50kN$

7 0

3 years ago

Read 2 more answers

6. What is the mass of a boy who is standing on top of a 1.5 meter high

puteri [66]

Answer:

m = 50 [kg]

Explanation:

The potential energy can be calculated by means of the following equation.

$E_{p}=m*g*h$

where:

Ep = potential energy = 735 [J]

m = mass [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 1.5 [m]

Now replacing:

$735=m*9.81*1.5\\m = 50 [kg]$

8 0

3 years ago

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab (Fig. 6-58). The coefficient of static fricti

vovikov84 [41]

Answer:

<u> $\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}$ </u>

Explanation:

Normal reaction from 40 kg slab on 10 kg block

M × g = 10 × 9.8 = 98 N

Static frictional force = 98 × 0.7 N

Static frictional force = 68.6 N is less than 100 N applied

10 kg block will slide on 40 kg slab and net force on it

= 100 N - kinetic friction

$=100-(98 \times 0.4)\left(\mu_{\text {kinetic }}=0.4\right)$

= 100 - 39.2

= 60.8 N

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with, } \frac{\mathrm{Net} \text { force }}{\text { mass }}$

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=\frac{60.8}{10}$

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=6.08 \mathrm{m} / \mathrm{s}^{2}$

$\text { Frictional force on 40 kg slab by 10 kg block, normal reaction \times \mu_{kinetic } }$

Frictional force on 40 kg slab by 10 kg block = 98 × 0.4

Frictional force on 40 kg slab by 10 kg block = 39.2 N

$40 \mathrm{kg} \text { slab will move with } \frac{\text { frictional force }}{\text { mass }}$

$40 \mathrm{kg} \text { slab will move with }=\frac{39.2}{40}$

40 kg slab will move with = $0.98 \mathrm{m} / \mathrm{s}^{2}$

$\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}$

3 0

3 years ago

A construction team gives an estimate of three months to repave a large stretch of a very busy road. The government responds tha

Keith_Richards [23]

Answer:

D) Three times as many workers are needed

Explanation:

Let the number of workers needed for this project be x. Then the total number of man-month needed for this project originally (3 months) is x * 3 = 3x man month. Now since the government wants to reduce the time length from 3 months to 1 month then the new number of workers needed is

3x / 1 = 3x

So 3 times the original amount. D) is the correct answer

8 0

4 years ago

Read 2 more answers