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2 years ago

Difference between 3.15 m and 2.0 m with the correct number of significant figures?

Physics

1 answer:

pentagon [3]2 years ago

3 0

Answer:

1.2

Explanation:

took the test

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Two ladders of uniform density and equal mass m are propped up against

-BARSIC- [3]

Answer:

i need some more coins LMAI

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3 years ago

A 75.0-kg ice skater moving at 10.0 m/s crashes into a stationary skater of equal mass. After the collision, the two skaters mov

Ksenya-84 [330]

Answer:

Explanation:

Momentum change for either skater is mΔv = 75.0(5.0) = 375 kg•m/s

As a change in momentum is equal to an impulse

375 = FΔt

F = 375/0.100 = 3750 N

As 3750 N < 4500 N no bones are broken.

4 0

2 years ago

If you weigh 882 N on EARTH (HINT: What number do we ALWAYS use for gravity on Earth), what is your mass?

Softa [21]

882 divided by 9.81 (this is acceleration due to gravity) it equals 89.91

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2 years ago

An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th

neonofarm [45]

Answer:

Electric force, $F=2.24\times 10^{-14}\ N$

Explanation:

It is given that,

Charge on an electron is $-1.6\times 10^{-19}\ C$

Electric field, $E=1.4\times 10^5\ N/m$

We need to find the magnitude of the electric force on this electron due to this field. The electric force is given by :

$F=qE\\\\F=1.6\times 10^{-19}\times 1.4\times 10^5\\\\F=2.24\times 10^{-14}\ N$

So, the electric force is $2.24\times 10^{-14}\ N$ .

6 0

2 years ago

A delivery truck travels with a constant velocity up an 8 slope. A 60 kg box sits on the floor of the truck and, because of sta

Blababa [14]

Explanation:

Given that,

The slope of the ramp, $\theta=8^{\circ}$

Mass of the box, m = 60 kg

(a) Distance covered by the truck up the slope, d = 300 m

Initially the truck moves with a constant velocity. We know that the net work done on the box is equal to 0 as per work energy theorem as :

$W=\dfrac{1}{2}m(v^2-u^2)=0$

u and v are the initial and the final velocity of the truck

(b) The work done on the box by the force of gravity is given by :

$W=Fd\ cos\theta$

Here, $\theta=90+8=98^{\circ}$

$W=mgd\ cos\theta$

$W=60\times 9.8\times 300\ cos(98)$

W = -24550.13 J

(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.

(d) The work done by friction is given by :

$W_f=-W$

$W_f=24550.13\ J$

Hence, this is the required solution.

6 0

3 years ago