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2 years ago

What is the mass of a large dog that weighs 441 newtons

1 answer:

5 0

Weight, w = mg. g ≈ 9.8 m/s². m = mass in kg. w is weight in N

441 N = m* 9.8

9.8m = 441

m = 441/9.8

m = 45 kg.

Mass of the dog is = 45 kg

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Given : Initial velocity = -1.3 m/s

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Time = 25 minutes.

To find : Average acceleration.

Solution: We are given units in meter/second (m/s).

So, we need to convert time 25 minutes in seconds.

1 minute = 60 seconds.

25 minutes = 60*25 = 1500 seconds.

Formula for average acceleration is given by,

$\frac{Final \ velocity -Initial \ velocity}{Final \ time - Initial \ time }$

We are not given intial time, so we can take initial time =0.

Plugging values in the above formula.

$Average \ acceleration = \frac{-6.5 -(-1.3)}{1500-0}$

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2 years ago

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Ryan and Becca are moving a folding table out of the sunlight. A cup of lemonade, with the message 0.44 kg is on the table. Becc

jarptica [38.1K]

Calculate the weight of the table through the equation,

W = mg

where W is the weight, m is the mass, and g is the acceleration due to gravity. Substituting the known values,
W = (0.44 kg)(9.8 m/s²)
W = 4.312 N

The components of this weight can be calculated through the equation,

Wx = W(sin θ)

and Wy = W(cos θ)

x - component:
Wx = W(sin θ)
Substituting,
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2 years ago

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2 years ago

In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is

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Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

$\theta$ = Angle = 20°

y = -20 cm

Velocity components

$u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s$

$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Acceleration components

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

Time taken is 0.26088 seconds

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m$

The distance the beetle travels on the ground is 0.3677181864 m

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