Find the mass of a flying discus that has a net force of 1.05 newtons and accelerates at 3.5 m/s^2

A runner completes a 10 km run in about 45 minutes. What was the runner’s speed in km/h?

OlgaM077 [116]

Answer: 12.5 km/s

I don't really know how to explain this, but here is your answer.

5 0

3 years ago

Help me quick!!! please!!

Rudik [331]

Answer:

39.240 W

Explanation:

Let's start by calculating the work done by the engine. We can assume that it is the same work done by the weight of the object to bring it from 40m to the surface: as much energy it takes to bring it up, the same ammount it takes to bring it down. Said work is $w= \vec F\cdot \vec{h} = mg h = 1000 \times 9.81\times 40 = 392.400 J$

At this point we can simply apply the definition of power, that is $P = \frac wt$ , to get the power of the engine is $39.240 W$

4 0

2 years ago

Read 2 more answers

A 1900kg car starts from rest and drives around a flat 65-m-diameter circular track. The forward force provided by the car's dri

s344n2d4d5 [400]

Answer:

$The\quad magnitude\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad \quad =2.52m/{ s }^{ 2 }\\ The\quad direction\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad =15.{ 72 }^{\o}\\The\quad car\quad begins\quad to\quad slide\quad out\quad \quad of\quad the\quad circle\quad after\quad 26.09s.\quad \quad \quad \quad$

Explanation:

8 0

3 years ago

A student initially 10.0 m East of his school walks 17.5 m West. The magnitude of the student's displacement, relative to the sc

Fantom [35]

Answer:

1. 7.5 m

2. towards west side

explanation:

I hope it will help you

3 0

2 years ago

The formula v = √2.5r models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radiu

mel-nik [20]

Answer:

The maximum safe speed of the car is 30.82 m/s.

Explanation:

It is given that,

The formula that models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r r, is in feet is given by :

$v=\sqrt{2.5\ r}$ .........(1)

A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet, r = 380 feet

Put the value of r in equation (1) as :

$v=\sqrt{2.5\ \times 380}$

v = 30.82 m/s

So, the maximum safe speed of the car is 30.82 m/s. Hence, this is the required solution.

4 0

3 years ago