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Natasha2012 [34]
3 years ago
15

4. It's important to know that science is:

Physics
2 answers:
Georgia [21]3 years ago
3 0

Answer:

<h2>What are we supposed to write about</h2>

Explanation:

EleoNora [17]3 years ago
3 0
It’s important to know that science is the future. Without science we would be dying from diseases and other illnesses.
You might be interested in
Why is the ozone layer important?
adoni [48]

Answer:

Ozone is a gas in the atmosphere that protects everything living on the Earth from harmful ultraviolet (UV) rays from the Sun. Without the layer of ozone in the atmosphere, it would be very difficult for anything to survive on the surface.

Explanation:

Ozone layer, also called ozonosphere, region of the upper atmosphere, between roughly 15 and 35 km (9 and 22 miles) above Earth’s surface, containing relatively high concentrations of ozone molecules (O3). Approximately 90 percent of the atmosphere’s ozone occurs in the stratosphere, the region extending from 10–18 km (6–11 miles) to approximately 50 km.

7 0
3 years ago
Read 2 more answers
A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo le
Alona [7]

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

R = u² sin2θ/g

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

u² = \frac{Rg}{sin2\theta}

u^2 = \frac{9.1 x 9.80}{sin26}

u^2 = 113.17

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

v^2 = (v_osin\theta)^2 -2gh

h =\frac{(v_o^2sin\theta)^2}{2g}

h  =  \frac{(113.17)(sin26)^2}{(2 x 9.80)}}

h = 1.10 m

7 0
3 years ago
Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially
AnnZ [28]

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

6 0
3 years ago
for any object suspended by any number of ropes, wires, or chains, how is the total amount of tension (tension in each rope adde
Sveta_85 [38]

Answer:

To calculate the tension on a rope holding 1 object, multiply the mass and gravitational acceleration of the object. If the object is experiencing any other acceleration, multiply that acceleration by the mass and add it to your first total.

Explanation:

The tension in a given strand of string or rope is a result of the forces pulling on the rope from either end. As a reminder, force = mass × acceleration. Assuming the rope is stretched tightly, any change in acceleration or mass in objects the rope is supporting will cause a change in tension in the rope. Don't forget the constant acceleration due to gravity - even if a system is at rest, its components are subject to this force. We can think of a tension in a given rope as T = (m × g) + (m × a), where "g" is the acceleration due to gravity of any objects the rope is supporting and "a" is any other acceleration on any objects the rope is supporting.[2]

For the purposes of most physics problems, we assume ideal strings - in other words, that our rope, cable, etc. is thin, massless, and can't be stretched or broken.

As an example, let's consider a system where a weight hangs from a wooden beam via a single rope (see picture). Neither the weight nor the rope are moving - the entire system is at rest. Because of this, we know that, for the weight to be held in equilibrium, the tension force must equal the force of gravity on the weight. In other words, Tension (Ft) = Force of gravity (Fg) = m × g.

Assuming a 10 kg weight, then, the tension force is 10 kg × 9.8 m/s2 = 98 Newtons.

7 0
3 years ago
Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, in
olya-2409 [2.1K]

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

I=\Delta p = m\Delta v

where

m = 62.0 kg is the mass of the passenger

\Delta v is the change in velocity of the car (and the passenger), which is

\Delta v = 5.30 m/s - 0 = 5.30 m/s

So, the linear impulse experienced by the passenger is

I=(62.0 kg)(5.30 m/s)=328.6 kg m/s

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

I=F \Delta t

where in this case

I=328.6 kg m/s is the linear impulse

\Delta t = 0.812 s is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N

7 0
3 years ago
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