How much force does an 88kg astronaut exert on his chair while accelerating straight up at 10 m/s^2

if a cat is running at a constant speed of 10km/h for 5 s, what is its average speed and what is its instantaneous speed at 4 s?

iris [78.8K]

Here a cat is running at constant speed which is given as 10 km/h for 5s

So here the average speed is defined as total distance moved in total time interval

so here it is given by

$v_{avg} = \frac{distance}{time}$

since

here speed of cat is constant so it will remain the same

And hence the average speed and instantaneous speed at any instant for this duration will remain the same

so here answer would be

<em>average speed = 10 km/h</em>

<em>instantaneous speed = 10 km/h</em>

8 0

3 years ago

Describe three societal needs that drive scientific endeavor and give an example of a scientific endeavor associated with each.

Trava [24]

The answer is:

1. Cure disease

2. Improve livelihoods

3. Understand nature and themselves.

1. Disease has plagued humans since the first man. Therefore, man has always used science to seek a cure for any emerging disease such as cancer today

2. Humans use science to improve livelihoods such as means of faster traveling and increasing yields in the fields

3. Science has been used to understand humans and the environment in which they live. This is evident with the numerous scientific probes that explore the earth and space collecting and analyzing data.

6 0

4 years ago

A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a

pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

$120^2=130^2+50^2-2\times130\times50\cos\alpha$

$\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}$

$\alpha=\cos^{-1}(0.3846)$

$\alpha=67.38^{\circ}$

We need to calculate the angle β

Using cosine law

$50^2=130^2+120^2-2\times130\times120\cos\beta$

$\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}$

$\beta=\cos^{-1}(0.923)$

$\beta=22.63^{\circ}$

We need to calculate the force on 130 cm side

Using formula of force

$F_{130}=ILB\sin\theta$

$F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0$

$F_{130}=0$

We need to calculate the force on 120 cm side

Using formula of force

$F_{120}=ILB\sin\beta$

$F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63$

$F_{120}=0.1385\ N$

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

$F_{50}=ILB\sin\alpha$

$F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38$

$F_{50}=0.1385\ N$

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

8 0

3 years ago

an object moving at 10. km/hr has a kinetic energy of 10. J. what is the kinetic energy of the same object if it is moving at 20

Schach [20]

Kinetic energy is related to velocity by:
KE = (1/2)mv^2

solve for mass m
10 = (1/2)m(10)^2
10 = (1/2)m(100)
10= 50m
10/50 = m
1/5 = m

at 20 km/hr

KE = (1/2)(1/5)(20)^2
KE = (1/10)(400)
KE = 40 J

5 0

3 years ago

What is the massof the largest ruby?

alexandr402 [8]

I think the answer is 2283g

4 0

4 years ago