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3 years ago

3 m/s north is an example of a(n) ____.

2 answers:

6 0

It is velocity. Velocity is a vector quantity. Vector quantity has both magnitude and direction. 3 m/s north has 3 m/s as magnitude and north as direction.

alexandr1967 [171]3 years ago

4 0

Rate of speed (3 m/s north is three miles per second north, so it's a rate of speed)

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A soap bubble has a diameter of 4mm .Cakculate the pressure inside it if the atmospheric pressure outside is 10^5Nm^2 . The surf

Kipish [7]

Look it up it will give the right anwser

3 0

3 years ago

Read 2 more answers

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

3 years ago

A block of mass 27.00 kg sits on a horizontal surface with, coefficient of kinetic

zhannawk [14.2K]

Answer:

The force is $F = 172 \ N$

Explanation:

From the question we are told that

The mass of the block is $m_b = 27.0 \ kg$

The coefficient of static friction is $\mu_s = 0.65$

The coefficient of kinetic friction is $\mu_k = 0.50$

The normal force acting on the block is

$N = m * g$

substituting values

$N = 27 * 9.8$

$N = 294.6 \ N$

Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as

$F_f = \mu_s * N$

substituting values

$F_f = 0.65 * 264.6$

$F_f = 172 \ N$

Now for this block to move the force require is equal to $F_f$ i.e

$F= F_f$

=> $F = 172 \ N$

5 0

3 years ago

List the following in order from largest to smallest:

LiRa [457]

Organisms, organs, tissues, cell ....

a more specific answer would be:
organism, organ system, organ, tissue, cell, atom

4 0

3 years ago

A resistor R1 is wired to a battery, then resistor R2 is added in series. Are (a) the potential difference across R1 and (b) the

tia_tia [17]

Answer:

Explanation:

a ) Earlier emf of cell applied on R₁ but now emf will be distributed among R₁ and R₂

Potential difference on R₁ will become less .

b ) Current is inversely proportional to resistance of the circuit. As resistance increases , current will be less . So current through R₁ will become less.

c )

When resistance is added in series , they are added up to obtain equivalent resistance . So equivalent resistance R₁₂ will be more than R₁ OR R₂.

6 0

3 years ago