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nasty-shy [4]
3 years ago
8

A 300.0-kg speedboat is moving across a lake at 35.0 m/s. what is the momentum of the speedboat?

Physics
2 answers:
LUCKY_DIMON [66]3 years ago
8 0

Answer: The correct answer is 10,500 kg m/s.

Explanation:

The expression of the momentum in terms of mass and velocity is as follows;

p=mv

Here, m is the mass of the object and v is the velocity.

It is given in the problem that a 300.0-kg speedboat is moving across a lake at 35.0 m/s.

Put m= 300 kg and v= 35 m/s in the above expression.

p=(300)(35)

p= 10,500 kg m/s

Therefore, the momentum of the speedboat is 10,500 kg m/s.

kozerog [31]3 years ago
4 0
We know that:
p=mv
p=300kg*35m/s
p=10500Ns
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The fire alarm goes off, and a 75 kg firefighter slides down a pole with a constant acceleration of a = 6 m/s square. What is th
Alex17521 [72]

Answer:

450N

Explanation:

Given data

Mass m= 75kg

Acceleration= 6m/s^2

From the Newtons first law, F=ma

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Hence the force is 450N

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How can you tell animal and plant cells apart
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8 0
3 years ago
What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an
astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution=T_b=101^{\circ} C

Boiling point water=100 degree Celsius

K_f=1.86K/m

K_b=0.512 K/m

\Delta T_b=T-T_0

Where T=Boiling point of solution

T_0=Boiling point of pure solvent

\Delta T_b=101-100=1^{\circ}C

\Delta T_b=k_bm

Using the formula

1=0.512\times m

Molality,m=\frac{1}{0.512} m

\Delta T_f=k_fm

Using the formula

\Delta T_f=\frac{1}{0.512}\times 1.86

\Delta T_f=3.63 C

We know that

\Delta T_f=T_0-T_1

Where T_0 =Freezing point of solvent

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Using the formula

3.63=0-T_1

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Hence, the freezing point of solution=-3.63 degree Celsius

7 0
3 years ago
UN BLOQUE TIENE DE DIMENSIONES AB = 4 cm, BC = 2.5 cm, BF = 6 cm Y UNA MASA DE 300 gr  DETERMINE SU VOLUMEN  HALLE SU DENSIDAD
torisob [31]

Responder:

- Volumen del bloque = 60cm³

- Densidad = 5g / cm³

Explicación:

Dada la dimensión del bloque

AB = 4 cm, BC = 2.5 cm y BF = 6 cm

Volumen del bloque = Largo × Ancho × Alto

Volumen del bloque = AB × BC × BF

Volumen del bloque = 4 × 2.5 × 6

Volumen del bloque = 60cm³

Dada la masa del bloque = 300 gramos

Densidad = Masa / Volumen

Densidad = 300g / 60cm³

Densidad = 5g / cm³

Algunos materiales que tienen una densidad de 5g / cm³ o cercanos son radio, germanio y europio con densidades de 5g / cm³, 5.3g / cm³ y 5.24g / cm³ respectivamente.

6 0
3 years ago
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