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3 years ago

The force of gravity acting upon an object is the objects mass. True or false

Physics

1 answer:

rosijanka [135]3 years ago

4 0

False , because mass is the amount of matter contains

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Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil

laila [671]

Answer:

$W_n_e_t=7.648512 \approx 7.6J$

$K.E=0.8J$

$v=0.7844645406 \approx 0.78m/s$

Explanation:

From the question we are told that

Mass of pitcher $M= 2.6kg$

Force on pitcher $f=8.8N$

Distance traveled $48cm=>0.48m$

Coefficient of friction $\mu=0.28$

a)Generally frictional force is mathematically given by

$F=\mu N$

$F=0.28*2.6*9.8$

$F=7.1344N$

Generally work done on the pitcher is mathematically given as

$W_n_e_t=W_f+W_F$

$W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N$

$W_n_e_t=4.224-3.424512$

$W_n_e_t=0.799488\approx 0.8J$

b)Generally K.E can be given mathematically as

$K.E= W_n_e_t$

Therefore

$K.E=0.8J$

c)Generally the equation for kinetic energy is mathematically represented by

$K.E=1/2mv^2$

$0.8=1/2mv^2$

Velocity as subject

$v=\sqrt{\frac{K.E*2}{m} }$

$v=\sqrt{\frac{0.8*2}{2.6} }$

$v=0.7844645406 \approx 0.78m/s$

6 0

3 years ago

The force of friction between an object and the surface upon which it is sliding is 14N and the coefficient of friction between

Nat2105 [25]

Answer:

14 x 0.27 = 3.78 is your answer

Explanation:

the question is asking for the weight of the object so you multiply and get 3.79

5 0

3 years ago

A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. How far away is the reflecting object? (c

Assoli18 [71]

A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. The reflecting object would be 3000 m away .

Phenomenon of hearing back our own sound is called an echo. It is due to successive reflection of sound waves from the surfaces or obstacles of large size. To hear an echo, there must be a time gap of 0.1 second in original sound and the reflected sound.

Given

time = 20 μs = 20 * $10^{-6}$ s

let distance to the reflecting surface be = x

total distance travelled by pulse will be = 2x

speed = 3.0 × $10^{8}$ m/s

distance = speed * time

2x = 3.0 × $10^{8}$ * 20 * $10^{-6}$

x = 3000 m

The reflecting object would be 3000 m away

To learn more about echo here

brainly.com/question/14861578?referrer=searchResults

#SPJ4

5 0

2 years ago

Martha was leaning out of the window on the second floor of her house and speaking to Steve. Suddenly, her glasses slipped from

DiKsa [7]

Answer:

s = 23.72 m

v = 21.56 m/s²

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

u = 0 m/s

g = 9.8 m/s²

s = 0 + 0.5 × 9.8 × 2.2²

s = 23.72 m

b) impact velocity

v = √(2gh)

v = √(2× 9.8 × 23.72)

v = √464.912

v = 21.56 m/s²

6 0

4 years ago

3 positive test charge lie on the x axis,q1=25microcoloum,is at the origin ,q2=10microcoloumb is at x=2m and q3=20microcloumb is

alexira [117]

Answer:

the link is right here

6 0

3 years ago