All categories

3 years ago

Which of the following could be units of speed? A. m B. m/s C. m north D. degrees north

2 answers:

34kurt3 years ago

7 0

Since velocity is vector quantity, it'll depend on both direction as well as magnitude hence options C and D are rejected.
Speed = distance/time
Therefore, unit of velocity is m/s.

patriot [66]3 years ago

4 0

B. M/s because north is to to with locating areas degrees north is temperature and speed is not measure in meters

You might be interested in

PLEASE ANSWER ASAP, WILL MARK BRAINLIEST***

Andreyy89

The answer is to your question is c

7 0

3 years ago

If two arm wrestlers exert a force on each other’s hands, and the hands don’t move, the forces must be *

jeyben [28]

Balanced. They’re equally as strong so as their arm wrestling, neither of the men’s hands go down. Because they’re equally/balanced as strong.

3 0

3 years ago

Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 77.3 N, Jill pull

DiKsa [7]

Answer:

$F_{net} = 232.8 N$

towards right so it is -15 degree

Explanation:

Net force in forward direction due to all three is given as

$F_x = F_1 + F_2cos45 + F_3cos45$

here we know that

$F_1 = 77.3 N$

$F_2 = 61.7 N$

$F_3 = 147 N$

$F_x = 77.3 + 61.7 cos45 + 147 cos45$

$F_x = 224.9 N$

Similarly in Y direction we will have

$F_y = F_3 sin45 - F_2 sin45$

$F_y = (147 - 61.7)sin45$

$F_y = 60.3 N$

Now the net force on the donkey is given as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$F_{net} = \sqrt{224.9^2 + 60.3^2}$

$F_{net} = 232.8 N$

Now direction of force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{60.3}{224.9}$

$\theta = 15^o$ towards right so it is -15 degree

3 0

3 years ago

a spring is compressed 15 centimeters when a 8.5 kilogram weight is set upon it. how much power would be needed to stretch out t

o-na [289]

Answer:

159.38 Watts

Explanation:

Initially;

Mass on the spring is 8.5 kg
Therefore, compression force is 85 N
Compression distance is 15 cm or 0.15 m

But;

F = kx

where F is the force of compression, k is the spring constant and x is the compression distance.

Thus;

k = F/x

= 85 N ÷0.15

= 566.67 N/m

We are required to determine the power needed to stretch the same spring for 1.5 m in 4 secs.

Power = Work done ÷ time

Work done is given by 0.5kx²

Therefore;

Power = 0.5kx²÷ t

= (0.5×566.67 N/m × 1.5² ) ÷ 4 seconds

= 159.38 Watts

Thus, the power needed is 159.38 watts

5 0

4 years ago

Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0

3 years ago