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2 years ago

7

A 25-kilogram space probe fell freely with an acceleration of 2.0 m/s2 just before it landed on a distant planet. What is the we

ight of the space probe on that planet?

1 answer:

malfutka [58]2 years ago

8 0

the weight of the space probe on that planet is 50 N

To calculate the weight of the space probe, we use the formula below.

Formula:

W = mg

Where:

W = weight of the space probe on the planet
m = mass of the space probe
g = acceleration due to gravity.

From the question,

Given:

m = 25 kg
a = 2 m/s

Substitute these values into equation 1

W = 25(2)
W = 50 N.

Hence, the weight of the space probe on that planet is 50 N

Learn more about weight here: brainly.com/question/229459

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What is the momentum of an object in splace

forsale [732]

Answer:

The total momentum of the universe is always the same and is equal to zero. The total momentum of an isolated system never changes. Momentum can be transferred from one body to another.

Momentum quantifies how likely an object is to stay in motion. Momentum can also be explained using the equation, p=mv, where p is equal to momentum, m is equal to mass, and v is equal to velocity.

Explanation:

5 0

3 years ago

A burglar attempts to drag a 108 kg metal safe across a polished wood floor Assume that the coefficient of static friction is 0.

V125BC [204]

Answer:

2.00 m/s²

Explanation:

Given

The Mass of the metal safe, M = 108kg

Pushing force applied by the burglar, F = 534 N

Co-efficient of kinetic friction, $\mu_k$ = 0.3

Now,

The force against the kinetic friction is given as:

$f = \mu_k N = u_k Mg$

Where,

N = Normal reaction

g= acceleration due to the gravity

Substituting the values in the above equation, we get

$f = 0.3\times108\times9.8$

or

$f = 317.52N$

Now, the net force on to the metal safe is

$F_{Net}= F-f$

Substituting the values in the equation we get

$F_{Net}= 534N-317.52N$

or

$F_{Net}= 216.48$

also,

$F_{Net}= M\times$ acceleration of the safe

Therefore, the acceleration of the metal safe will be

acceleration of the safe= $\frac{F_{Net}}{M}$

or

acceleration of the safe= $\frac{216.48}{108}$

or

acceleration of the safe= $2.00 m/s^2$

Hence, the acceleration of the metal safe will be 2.00 m/s²

3 0

2 years ago

A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hang

valina [46]

Answer:

work done is -2.8 × 10⁻⁶ J

Explanation:

Given the data in the question;

mass of the pendulum m = 6 kg

Length of core = 1.7 m

Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad

In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.

Now, the required work done will be;

$W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta$

$W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta$

$W = -mgl \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta$

W = $-mgl[$ -cosθ $]^{0.047}_{0.045 }$

W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]

W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]

W = -2.8 × 10⁻⁶ J

Therefore, work done is -2.8 × 10⁻⁶ J

6 0

3 years ago

An asteroid has a mean radius equal to 17 times that of Earth. Using Kepler's 3rd law find its period. T^2=a^3

vagabundo [1.1K]

Answer:

If T^2 = a^3

T1^2 / T2^2 = a1^3 / a2^3

Or T2^2 = T1^2 * ( a2^3 / a1^3)

Given T1 = 1 yr and a2 / a1 = 17

Then T2^2 = 1 * 17^3

or T2 = 70 yrs

8 0

2 years ago

A block is sliding down an inclined plane that makes an angle of 65o with the horizontal. If the coefficient of kinetic friction

Nezavi [6.7K]

Answer:

8.2 m/s²

Explanation:

m = mass of the block

μ = Coefficient of kinetic friction = 0.17

$F_{n}$ = Normal force on the block by the ramp

$f_{k}$ = kinetic frictional force

Force equation perpendicular to ramp surface is given as

$F_{n} = mg Cos65$

Kinetic frictional force is given as

$f_{k} = \mu F_{n}$

$f_{k} = \mu mg Cos65$

Force equation parallel to ramp surface is given as

$mg Sin65 - f_{k} = ma$

$mg Sin65 - \mu mg Cos65 = ma$

$g Sin65 - \mu g Cos65 = a$

$(9.8) Sin65 - (0.17) (9.8) Cos65 = a$

$a = 8.2$ m/s²

4 0

3 years ago