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3 years ago

How much force is required to accelerate a 5kg mass pushed by a 10n force

Physics

1 answer:

Rama09 [41]3 years ago

4 0

2 m/s^2

^2 means squared

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Bob rides his bike with a constant speed of 10 miles per hour. How long will he take to travel a distance of 15 miles?

ICE Princess25 [194]

${\underline{\pink{\textsf{\textbf{ Answer : }}}}}$

➡ 150hrs.

${\underline{\pink{\textsf{\textbf{Explanation : }}}}}$

➡ Time = distance × speed

➡ Time = 15*10

➡ Time = 150hrs ans.

7 0

2 years ago

A 2.3 kg particle-like object moves in a plane with velocity components vx = 40 m/s and vy = 75 m/s as it passes through the poi

Sonbull [250]

Answer:

(a) $\overrightarrow{L}=885.5\widehat{k}$

(b) $\overrightarrow{L}=1046.5\widehat{k}$

Explanation:

mass, m = 2.3 kg

vx = 40 m/s

vy = 75 m/s

(a) Angular momentum is given by

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

Where, p is the linear momentum and r is the position vector about which the angular momentum is calculated.

Here, $\overrightarrow{r}=3\widehat{i}-4\widehat{j}$

$\overrightarrow{p}=m\overrightarrow{v}$

$\overrightarrow{p}=2.3\left ( 40\widehat{i}+75\widehat{j} \right )$

$\overrightarrow{p}= 92\widehat{i}+172.5\widehat{j}$

So, the angular momentum

$\overrightarrow{L}=\left ( 3\widehat{i}-4\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )$

$\overrightarrow{L}=885.5\widehat{k}$

(b) Here, $\overrightarrow{r}=(3+2)\widehat{i}+(-4+2)\widehat{j}$

$\overrightarrow{r}=5\widehat{i}-2\widehat{j}$

$\overrightarrow{L}=\left ( 5\widehat{i}-2\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )$

$\overrightarrow{L}=1046.5\widehat{k}$

6 0

2 years ago

The color of a star depends on its , and -colored stars are the coolest.

wlad13 [49]

Answer:

temperature and red

is the answer

Explanation:

7 0

2 years ago

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Work is done on an object only if the force and displacement are __________?

miskamm [114]

It’s either movement or work or that’s what a quizzie said

8 0

2 years ago

Use the following equation to help you answer the question. The peak intensity of radiation from a star named Sigma is 2 x 10 6

puteri [66]

Answer:

1449 K

Explanation:

The surface temperature of a star is related to its peak wavelength by Wien's displacement law:

$T=\frac{b}{\lambda}$

where

T is the surface temperature

b is Wien's displacement constant

$\lambda=2\cdot 10^{-6} m$

So the surface temperature of the star is

$T=\frac{2.898 \cdot 10^{-3} Km}{2\cdot 10^{-6} m}=1449 K$

8 0

2 years ago

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