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3 years ago

Bruce has a momentum of 430kg m/s and is running at 7.8m/s. What is Bruce's mass?

Physics

2 answers:

mixas84 [53]3 years ago

8 0

Answer:

55.13 kg

Explanation:

Momentum, P = m*v

P = m*v

m = P/v

m = 430/7.8

m = 55.13 kg

ryzh [129]3 years ago

7 0

Answer:

55.128kg

Explanation:

P= m×v

or, m=P/v = 430/7.8 = 55.128 kg

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What scientists do that is the basis of their investigations

matrenka [14]

<span>The "scientific method" consists of forming a hypothesis (a statement about what the researchers think is true), collecting data that might confirm or refute the hypothesis, and then examining the data to see how well or poorly they support the original hypothesis.</span>

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4 years ago

Rank the deformations of the following rods in terms of the magnitude of the average normal strain: (a) The length of a 1-m-long

kipiarov [429]

To solve this problem we will consider the concepts related to the normal deformation on a surface, generated when the change in length is taken per unit of established length, that is, the division between the longitudinal fraction gained or lost, over the initial length. In general mode this normal deformation can be defined as

$\epsilon = \frac{\delta}{l} = \frac{l_0-l}{l}$

Here,

$\delta$ = Change in final length $(l_0)$ and the initial length $l$

PART A)

$\epsilon = \frac{\delta_1}{l}$

$\epsilon = \frac{l_0-l}{l_0}$

$\epsilon = \frac{1.02-1}{1}$

$\epsilon = 0.01961$

PART B)

$\epsilon = \frac{\delta_1}{l}$

$\epsilon = \frac{l_0-l}{l_0}$

$\epsilon = \frac{2-1.05}{2}$

$\epsilon = 0.475$

PART C)

$\epsilon = \frac{\delta_1}{l}$

$\epsilon = \frac{l_0-l}{l_0}$

$\epsilon = \frac{3.07-3}{3}$

$\epsilon = 0.0233$

Therefore the rank of this deformation would be B>C>A

7 0

3 years ago

The planet Uranus has a radius of 25,360 km and a surface acceleration due to gravity of 9.0 m/s^2 at its poles. Its moon Mirand

AlexFokin [52]

Answer:

$8.67791\times 10^{25}\ kg$

$0.34589\ m/s^2$

$0.07903\ m/s^2$

Explanation:

M = Mass of Uranus

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Uranus = 25360 km

h = Altitude = 104000 km

$r_m$ = Radius of Miranda = 236 km

m = Mass of Miranda = $6.6\times 10^{19}\ kg$

Acceleration due to gravity is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow M=\dfrac{gr^2}{G}\\\Rightarrow M=\dfrac{9\times 25360000^2}{6.67\times 10^{-11}}\\\Rightarrow M=8.67791\times 10^{25}\ kg$

The mass of Uranus is $8.67791\times 10^{25}\ kg$

Acceleration is given by

$a_m=\dfrac{GM}{(r+h)^2}\\\Rightarrow a_m=\dfrac{6.67\times 10^{-11}\times 8.67791\times 10^{25}}{(25360000+104000000)^2}\\\Rightarrow a_m=0.34589\ m/s^2$

Miranda's acceleration due to its orbital motion about Uranus is $0.34589\ m/s^2$

On Miranda

$g_m=\dfrac{Gm}{r_m^2}\\\Rightarrow g_m=\dfrac{6.67\times 10^{-11}\times 6.6\times 10^{19}}{236000^2}\\\Rightarrow g_m=0.07903\ m/s^2$

Acceleration due to Miranda's gravity at the surface of Miranda is $0.07903\ m/s^2$

No, both the objects will fall towards Uranus. Also, they are not stationary.

6 0

3 years ago

A 58.1 kg wood board is resting on very smooth ice in the middle of a frozen lake. A 48.3 kg boy stands at one end of the board.

VLD [36.1K]

Answer:

V = 1.41 m /s

Explanation:

Movement in both wood board and boy is created by internal forces . No horizontal external force acts on them . So conservation of momentum will apply on both of them for movement in horizontal direction.

Mass of wood board x velocity = mass of boy x velocity

58.1 x V = 48.3 x 1.69

V = 1.41 m /s

4 0

4 years ago

Do you think it would take more force to transport five blocks in the sled or in the wagon?

enot [183]

It will depend on the frictional force involved in the two. I think it will take more force in sled.

6 0

4 years ago

Read 2 more answers