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4 years ago

Change in speed over a given period of time is

Physics

2 answers:

Dvinal [7]4 years ago

8 0

Explanation:

Acceleration is the change in speed over a given time period

Maurinko [17]4 years ago

6 0

Rate. because, ANY change over time is called a rate. acceleration is the rate of change of velocity. so your answer is Rate.

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Compare the kinetic energies of a ball moving at 5 m/s to when it is moving

Fantom [35]

Answer:

Explanation:

KE = mv²/2

The more velocity (speed) is, the more kinetic energy an object has.

Because 10 m/s > 5 m/s, KE at 10 m/s > KE at 5 m/s.

5 0

3 years ago

Joy uses 20n of force to shovel the snow 10 meters. how much work does she do? 200 j 2 j 30 j 100 j

levacccp [35]

20 newtons divided by 10 meters

8 0

3 years ago

An object in free fall travels a distance s that is directly proportional to the square of the time t. If an object falls 1088 f

ikadub [295]

Answer:

1,700feet

Explanation:

If an object in free fall travels a distance s that is directly proportional to the square of the time t, this can be represented mathematically as;

S = kt²where;

k is the proportionality constant

K = s/t²

s1/t1²= s2/t2²= Sn/tn²= k for values of the distance and time. Using the formula

s1/t1² = s2/t2² where;

s1 is the falling distance in time t1 s2 is the falling distance in time t2

Given s1 = 1088feet, t1 = 8secs, s2 = ? t2 = 10secs

Substituting this value in the formula to get s2, we have;

1088/8²= s2/10²

64s2= 108800

s2 = 108800/64

s2 = 1,700feet

This means the object will fall a distance of 1,700feet in 10seconds

5 0

3 years ago

What is the mass of a crane that has a weight of 697,005.40N?

igomit [66]

Explanation:

hope this helps you out if not im sorry

6 0

3 years ago

A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140

max2010maxim [7]

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

$u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\$

Using formula:

$v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\$

$= 0.55 \ \frac{m}{sec^2}\\\\$

$F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\$

Calculating the Work by net force

$W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\$

The above work is converted into thermal energy.

Now,

$F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J$

6 0

3 years ago