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blagie [28]
2 years ago
9

I need help with this problem

Physics
1 answer:
frosja888 [35]2 years ago
6 0

Answer:

v = dr/dt = 7t x + 0 y + (- 4) z

a = dv/dt = 7 x + 0 y + 0 z

t = 0:

v = (-4) z

a = 7 x

Explanation:

v = dr/dt = 7t x + 0 y + (- 4) z

a = dv/dt = 7 x + 0 y + 0 z

t = 0:

v = (-4) z

a = 7 x

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The first law states that “objects at rest and objects in motion remain in motion in a straight line unless acted upon by an unbalanced force”. Keeping the ice smooth will make sure there is not friction, friction would slow the puck down
5 0
3 years ago
Read 2 more answers
An engine is used to lift a beam weighing 9,800 N up to 145 m. How much work must the engine do to lift this beam? How much work
Arturiano [62]

Explanation:

Given that,

Weight of the engine used to lift a beam, W = 9800 N

Distance, d = 145 m

Work done by the engine to lift the beam is given by :

W = F d

W=9800\ N\times 145\ m\\\\W=1421000\ J\\\\W=1421\ kJ

Let W' is the work must be done to lift it 290 m. It is given by :

W'=9800\ N\times 290\ m\\\\W'=2842000\ J\\\\W'=2842\ kJ

Hence, this is the required solution.

5 0
4 years ago
A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you
soldi70 [24.7K]

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}       (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s

The interaction time to avoid that the water balloon breaks is 0.029s

5 0
4 years ago
Ana walks from 4 m to 200 cm. Which of the following statements is true about
Gemiola [76]
Distance=2m

because 200cm = 2m
so 4m-2m=2m
3 0
3 years ago
The speed of sound in air is approximately 340 m/s. The speed of light in air is approximately 3 x 108 m/s. If 10 seconds elapse
olya-2409 [2.1K]

Answer:

3400 m

Explanation:

Both lightning and thunder happen at the same time but one is faster than the other. The distance traveled by a sound can be calculated from its speed such that;

 speed = distance/time, hence, distance = speed x time.

<em>For a thunder with 340 m/s speed and 10 seconds away from lightning, the distance between the thunder and the lightning can be calculated as</em>;

distance = 340 m/s x 10 s = 3400 m

     

3 0
3 years ago
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