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NeTakaya
2 years ago
10

Which of the following is NOT a vector quantity?

Physics
1 answer:
lora16 [44]2 years ago
5 0

Answer:

D

Explanation:

Acceleration has both magnitude and direction. because velocity is vector quantity

average velocity is vector

linear momentum = mass× velocity

therefore it's also vector

force = mass× acceleration

acceleration is vector therefore force also vector

potential energy= mgh

energy in any form is scaler quantity therefore potential energy too is a scaler quantity.

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7 0
2 years ago
A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the
DiKsa [7]

Answer:

a). M = 20.392 kg

b). am = 0.56 m/s^2 (block),  aM = 0.28 m/s^2 (bucket)

Explanation:

a). We got  N = mg cos θ,

                  f = $\mu_s N$

                    = $\mu_s mg \cos \theta$

If the block is ready to slide,

T = mg sin θ + f

T = mg sin θ + $\mu_s mg \cos \theta$   .....(i)

2T = Mg ..........(ii)

Putting (ii) in (i), we get

$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$

$M=2(m \sin \theta + \mu_s mg \cos \theta)$

$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$

M = 20.392 kg

b). $(h-x_m)+(h-x_M)+(h'+x_M)=l$  .............(iii)

   Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

$-\ddot{x}-2\ddot x_M=0$

$\ddot x_M=\frac{\ddot x_m}{2}$

$a_M=\frac{a_m}{2}$   .....................(iv)

We got,   N = mg cos  θ

                $f_K=\mu_K mg \cos \theta$

∴ $T-(mg \sin \theta + f_K) = ma_m$

  $T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$  ................(v)

Mg - 2T = Ma_M

$Mg-Ma_M=2T$

$Mg-\frac{Ma_M}{2} = 2T$    (from equation (iv))

$\frac{Mg}{2}-\frac{Ma_M}{4}=T$   .....................(vi)

Putting (vi) in equation (v),

$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$

$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$

$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$

$a_m= 0.56 \ m/s^2$

Using equation (iv), we get,

a_M= 0.28 \ m/s^2

6 0
2 years ago
Can someone please help me with a physics assignment on Friday?
zalisa [80]

Yes what do you need help on

6 0
2 years ago
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A spring has a spring constant of 450 N/m. How much must this spring be stretched to store 49 J of potential energy?
Mazyrski [523]

Answer:

x = 0.47 m

Explanation:

PS = ½kx²

x = √(2PS/k) = √(2(49)/450)) = 0.466666...

6 0
2 years ago
How much work can a 22kw car engine do in 60 seconds
Sindrei [870]

Answer:

13.2 x 10^5 J

Explanation:

Power = work/time

Given

Power = 22kw

= 22 x 1000 = 22000w

Time = 60 secs

Therefore

22000 = work/60

Cross multiply

Work = 22000 x 60

= 1320000

= 13.2 x 10^5J

6 0
2 years ago
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