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tatiyna
2 years ago
5

A floating ice block is pushed through a displacement along a straight embankment by rushing water, which exerts a force on the

block. How much work does the force do on the block during the displacement
Physics
1 answer:
lions [1.4K]2 years ago
8 0

The question is incomplete. Here is the complete question.

A floating ice block is pushed through a displacement vector d = (15m)i - (12m)j along a straight embankment by rushing water, which exerts a force vector F = (210N)i - (150N)j on the block. How much work does the force do on the block during displacement?

Answer: W = 4950J

Explanation: <u>Work</u> (W), in physics, is done when a force acts on an object that has a displacement form a place to another:

W = F · d

As the formula shows, Work is a scalar product, i.e, it results in a number, so, Work only has magnitude.

Force and displacement for the ice block are in 2 dimensions, then work will be:

W = (210)i - (150)j · (15)i - (12)j

W = (210*15) + (150*12)

W = 3150 + 1800

W = 4950J

During the displacement, the ice block has a work of 4950J

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Olin [163]

Answer:

Speed is a "scalar" quantity

(C) is the correct answer

An object could travel at 10 m/s to some point and then return to the origin at 10 m/s for an average speed of 10 m/s, however it's displacement over that time would be zero for a net velocity of zero.

4 0
2 years ago
6. A 0.09 kg arrow hits a target at 22 m/s and penetrates 4 cm before stopping.
tatyana61 [14]
A ) v = v o - a t
0 = 22 - a · t
a · t = 22
d = v o · t - a t²/2
0.04 = 22 t - 22 t / 2
0.04 = 11 t
t = 0.04 : 11 = 0.003636 s
a = 22 / t
a = 6050 m/s²
F = m · a = 0.09 kg · 6050 m/s²
F ( target→arrow) = - 544.5 N
b ) F ( arrow→target ) = 544.5 N
c )  If the speed was doubled:  v = 44 m/s;
F = a m
a = 6050 m/s²
a · t = 44
t = 6050 : 0.04
t = 0.007272 s
d = 44 t - 44 t/2 = 22 t
d = 22 · 0.007272
d = 0.16 m = 16 cm
8 0
3 years ago
What force is required to accelerated a body with a mass of 15 kilograms at a rate of 8 m/s2
rjkz [21]

Force  =  (mass) · (acceleration)

           = (15 kg) · (8 m/s²)

           =  120 kg-m/s²  =  120 newtons
7 0
3 years ago
Rearrange the kinetic energy equation so that velocity is the subject
mars1129 [50]

Answer: V = sqrt(2K.E/M)

Explanation:

K.E = 1 /2MV^2

MV^2 = 2 K.E

V^2 = 2K.E /M

V = sqrt(2K.E/M)

6 0
3 years ago
A tiger leaps with a horizontal speed of 4.5 m/s from a boulder and lands 15 meters away.What is the vertical velocity with whic
Gelneren [198K]

Answer:

v_oy = 16.33 m/s

Explanation:

To find the vertical velocity of the tiger, you use the information about the horizontal velocity and maximum horizontal distance traveled.

You use the following formula for the range of the trajectory:

x_{max}=\frac{2v_{ox}v_{oy}}{g}     ( 1 )

v_ox: horizontal initial velocity = 4.5m/s

v_oy: vertical initial velocity = ?

g: gravitational acceleration = 9.8m/s^2

x_max: range of the trajectory = 15 m

You do v_oy the subject of the formula ( 1 ) and you replace the values of the other parameters in order to calculate v_oy:

v_{oy}=\frac{gx_{max}}{2v_{ox}}=\frac{(9.8m/s^2)(15m)}{2(4.5m/s)}\\\\v_{oy}=16.33\frac{m}{s}

hence, the initial vertical velocity of the tiger is 16.33m/s

8 0
3 years ago
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