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evablogger [386]

2 years ago

13

If mass is measured in kg and acceleration is measured in m/s^2, what units would force be measured in? This unit is also know a

s a newton (N).

1 answer:

Vlada [557]2 years ago

5 0

Mass=kg
Acceleration=m/s^2

Using Newtons second law

$\\ \sf\longmapsto Force=Mass\times Acceleration$

$\\ \sf\longmapsto Force=kgm/s^2$

$\\ \sf\longmapsto Force=Newton(N)$

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Examples were friction is not useful

expeople1 [14]

The old electricity meters with disk using sapphire bearings for friction to be minimal to not aggravate metering

7 0

3 years ago

Pls someone help 10points!!!

GREYUIT [131]

Answer:

it is 45

Explanation:

3 0

3 years ago

A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he rele

kirill115 [55]

Explanation:

It is given that,

Speed of the ball, v = 10 m/s

Initial position of ball above ground, h = 20 m

(a) Let H is the maximum height reached by the ball. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh'$

$h'=\dfrac{v^2}{2g}$

$h'=\dfrac{(10)^2}{2\times 9.8}$

h' = 5.1 m

The maximum height above ground,

H = 5.1 + 20

H = 25.1 meters

So, the maximum height reached by the ball is 25.1 meters.

(b) The ball's speed as it passes the window on its way down is same as the initial speed i.e. 10 m/s.

Hence, this is the required solution.

6 0

3 years ago

A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the

Ierofanga [76]

Answer:

The discharge rate is $Q = 0.0192 \ m^3 /s$

Explanation:

From the question we are told that

The diameter is $d = 60 \ mm = 0.06 \ m$

The head is $h = 6 \ m$

The coefficient of contraction is $Cc = 0.68$

The coefficient of velocity is $Cv = 0.92$

The radius is mathematically evaluated as

$r = \frac{d}{2}$

substituting values

$r = \frac{ 0.06 }{2}$

$r = 0.03 \ m$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.03)^2$

$A = 0.00283 \ m^2$

The discharge rate is mathematically represented as

$Q = Cv *Cc * A * \sqrt{ 2 * g * h}$

substituting values

$Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}$

$Q = 0.0192 \ m^3 /s$

6 0

2 years ago

1. Two blocks travel along a level frictionless surface. Block A is initially moving to the right at 5.0 m/s, while block B is i

ad-work [718]

Answer: 2.67 m/s

Explanation:

Given

Mass of block A is $m_a=2\ kg$

mass of block B is $m_b=3\ kg$

The initial velocity of block A $u_a=5\ m/s$

the initial velocity of block B is $u_b=0$

After collision velocity of block A is $v_a=1\ m/s$

Conserving momentum

$m_au_a+m_bu_b=m_av_a+m_bv_b\\\\2\times 5+3\times0=2\times 1+3\times v_b\\\\v_b=\dfrac{8}{3}=2.67\ m/s$

The momentum of block A after the collision is $P_a=2\times 1=2\ kg.m/s$

Therefore, there is no change in sign.

6 0

2 years ago