Ask question

Ask question

All categories

3 years ago

10

How much time does it take a rhino running at 40 m/s to run 250 m?

1 answer:

lions [1.4K]3 years ago

8 0

Answer:

Rhinoceroses are odd-toed ungulates native to sub-Saharan Africa and southern Asia, though all five living species have hugely contracted in range and number due to the influence of humans. Despite their titanic, tank-like bulk, rhinos can be amazingly swift: The fastest may reach at least 50 kilometers per hour (31 mph).

Explanation:

sana makatulong

You might be interested in

How will you demonstrate Newton's 1st Law of Motion at home? Brainstorm 2 or 3 possible demonstrations

Archy [21]

1. Roll a ball across a table into an object
2. Drop something

5 0

2 years ago

Help!!!! A loop of area 0.100 m^2 is oriented at 15.5 degree angle to a 0.500 T magnetic field. It rotates until it is at a 45.0

Nat2105 [25]

Answer:

Resulting Change in the Magnetic Flux $=0.013\:Wb$

Explanation:

Thanks!

3 0

3 years ago

An 87.6 g lead ball is dropped from rest from a height of 7.00 m. The collision between the ball and the ground is totally inela

bija089 [108]

Taking specific heat of lead as 0.128 J/gK = c

We have energy of ball at 7.00 meter height = mgh = $87.6*10^{-3}*9.81*7$

When leads gets heated by a temperature ΔT energy needed = mcΔT

= $87.6*10^{-3}*0.128*10^3$ ΔT

Comparing both the equations

$87.6*10^{-3}*9.81*7$ = $87.6*10^{-3}*0.128*10^3$ ΔT

ΔT = 0.536 K

Change in temperature same in degree and kelvin scale

So ΔT = 0.536 $^0C$

7 0

2 years ago

2. Unless an object at rest is acted upon by a force, it stays at rest due to it's __________.

Leokris [45]

Inertia. (Newton's Laws of Motion)

5 0

3 years ago

For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103

Alona [7]

Answer:

a) P = 44850 N

b) $\delta l =0.254\ mm$

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

$\sigma=\frac{\textup{Load}}{\textup{Area}}$

on substituting the values, we get

$345\times10^6=\frac{\textup{Load}}{0.00013}$

or

Load, P = 44850 N

Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>

b)The deformation ( $\delta l$ ) due to an axial load is given as:

$\delta l =\frac{PL}{AE}$

on substituting the values, we get

$\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}$

or

$\delta l =0.254\ mm$

3 0

2 years ago